#help-27
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Yeah
Like you're saying my junk in the middle is all of my alpha/2, or am I wrong?
I'm looking at the tutorial and they didn't really explain how they got their columns
Each row is my z-score, then, right?
I don't understand the first row, no
And they don't go over that in the tutorial
They were like "we're looking up this row", but there's not an explanation for how they chose that column
"Example 2
The height of plants in a certain garden are normally distributed with a mean of μ = 26.5 inches and a standard deviation of σ = 2.5 inches. Approximately what percentage of plants are greater than 26 inches tall?
Step 1: Find the z-score.
First, we will find the z-score associated with a height of 26 inches.
z-score = (x – μ) / σ = (26 – 26.5) / 2.5 = -0.5 / 2.5 = -0.2
Step 2: Use the z-table to find the percentage that corresponds to the z-score.
Next, we will look up the value -0.2 in the z-table:"
To me they don't explain the column thing at all
-0.2 = -0.20
Oohh
Wait
So for my z value here, I have 0.025, so you're saying I'll find that value at where my first decimal place is 1.9 and my second decimal place is 0.06 to get 1.96 when you add them?
Yes 1.9+0.06=1.96
1.9 being from the first column
And 0.06 is from the first row
Then 0.025 is the area to the right
So 1-0.025 is the area to the left
Got it! Fuck they should seriously just put a legend on there
That was the biggest reverse engineering of the century
Thank you so much
That makes so much sense
Glad to help. We've all been where you are
Yeah, I mean I'm a junior so it's not like I don't know how to study. I'm just stressed with finals and work and exams and projects. Everything is due all at once
Thank you
How do I give you karma on here?
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Let q be a definite quadratic form, and I want to show that it is either positive definite or negative definite.
\ I then assume that q is neither positive nor negative definite, which leads to a contradiction. So $\exists x \neq 0, q(x)>0$ and $y \neq 0$ such that $q(y)<0$.
\ Let's posit $\varphi: t \mapsto q(tx+(1-t)y)$ with $x$ and $y$ well chosen.
Thus $\varphi(t)=t^2 q(x)+(1-t)^2 q(y)+2t(1-t)b(x,y)$ where b is the polar form of q.
\
We have $\varphi(0)=q(y)<0$ and $\varphi(1)=q(x)>0$, $\varphi$ being continuous we deduce by the intermediate value theorem that $\exists t_0 \in (0,1) , \varphi(t_0)=0$. q being defined, it is necessary that $t_0 x+(1-t_0)y=0$, and here I'm stuck...
if x and y are collinear, $\exists \lambda, y=\lambda x$ and $q(y)=\lambda^2 q(x)<0$ which is not possible because $q(x)>0$
tm
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hey guys, this limits goes to 0, right?
wolfram gave me an error
,w lim n to inf (e)^n * n! / n^n
but, if I'm right, I think that, using the alternate series test, this goes to 0
,w lim n to inf (e)^n * n! / n^{n+1/2}
@ancient trellis Has your question been resolved?
hey guys, if I use n-th term test with x=e, the limit goes to inf
then, the series divergers for x=e
by comparison, it also happens in x=-e
so the radius of converge is (-e,e)
yeah, got it
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yes
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what's the context here, looks like they are expecting you to compute a numerical approximation?
yes but i assume this means you're writing a computer program to calculate it?
would help if there was more context to the problem like bungo asked
@wintry umbra Has your question been resolved?
like what topics/methods you've learned recently for this problem
you learned all of those just recently?
@wintry umbra Has your question been resolved?
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how
do you know how area and volume scale with length?
if i have a square and i make each side twice as long, how much bigger will the area be?
if i have a cube and i make each side twice as long, how much bigger will the volume be?
4 times larger?
correct
what about the volume question
and volume becomes 8× larger
no
if the length increases by n
the area goes up by n² and the volume by n³
this works for any shape
eg a circle's area is pi r²
if u double the radius u get pi (2r)² = 4 pi r²
so if the length of the statue went from 12 to 15
what factor did it increase by?
and the volume (and therefore the mass) will increase by that factor cubed
@stuck vortex Has your question been resolved?
@stuck vortex Do you still need help?
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im stuck on (j) i dont see any patterns, any hints?
normally when i want to prove an integer is good, i split it into a good integer and a bad integer, then try to prove that they combined is good (ok this doesent make sense lemme write an example)
M=30=5×6
by CRT, n^n=1 mod 5 and n^n=1 mod 6, since M=6 is good then n^n=1 mod 6 => n=1 mod 6, trying out numbers for n^n=1 mod 5 the ones that dont imply n=1 mod 5 are all in the form 4+20k, 8+20k, and 12+20k for all integers k, and these are all even which share a factor with 6 so if you choose one of them to be n, then its impossible for n^n to be 1 mod 6, so the only other possibility is that it implies n=1 mod 5 so M=30 is good
for proving bad i try to find a contradiction
@solid osprey Has your question been resolved?
9 should work
,w (9^9) mod 14
oh lemme look
smh how did i even mess that up
youre familiar with fermats little theorem right?
im gonna start spitballing while i figure this out
yea
ok, so lets think about totients then
whats special about totients of all odds up to 14 that 14 breaks?
i conjecture that a good M cant have a prime factor>3
all evens whoops
step d shows you 10 is good
honestly no idea
yes
ok lets look at that a little more
i just checked 3 5 9 and it worked for 9
by flt if a^b = 1 mod 14, we can reduce a mod 14 and b mod 6
mhm
i think because of that it cycles trough every odd a and b and prolly one of them is bound to be 1 mod 14?
i kinda half wrote each step but its really terribly written
Is this occupied
so in order to have an odd number, we must have ord of it be 3
an odd number with an odd ord
proof
odd coincidence
no pun intended
to have an odd number to be good?
to have an odd number disprove an even number from being good
oh
why do you need the ord of it to be 3? are you taking M=14 for the example and 3|6?
if you want n^n congruent to 1, you need n odd
since we reduce the power by an even number, we need it to be congruent to 3 mod 6
or 1
but a power of 1 would require a 1 as the base
ok and 5 mod 6 wont work because the ord divides 6 and the ord also has to divide 5 mod 6, and the gcd of them both is 1 but the ord cant be 1 so it has to be 3 mod 6 right?
yes
and i was looking through cubes that happen to work and 9^3 worked
so that was kinda lucky
i dont really know where to go
hm
so my thought process was if phi(M) was a power of two this cant happen
but then M=42 is fine
oh
thats because 3|42 and phi(42)=12
so we must have a power of 3 so that we have an odd order
you mean like the order has to be 3 for it to be an odd order?
but then x^3 = 1 mod 42 cant happen if x is a multiple of 3
well if the odd prime factors of phi(M) cant divide M then M is bad right?
thats what it looks like
but is that what it wants us to find tho cause seems hard to use
i dont know, but this feels like a step in the right direction
it does, maybe we can change it to be something more general and easier to put in use?
maybe thats enough for what we need if 2013 acts nice enough
because part k narrows down our search to 30 possibilities
,w factor 2013
totient(2013)=1200
yeah so (k) is saying if you add a factor of 2 and 5 (aka multiply by 10) then its good
this might just be prime factors in general
2 is a little touchy
ok yeah this looks good rn
oh yea, M is always divisible by 2 since M is even
but multiplying by 10 changes phi
like m=10k for k from 1 to 30, phi(2013m)=phi(20130k)=
phi(20130)phi(k)gcd(20130,k)/phi(gcd(20130,k)) only changes when k is either a multiple of 3 or 11
actually nvm
i think at this point, you should just test each case
some are pretty easy like 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 150, 160, 180, 200, 210, 220, 240, 250, 280, 300
thats a good majority of them
if it turns out this was for an aime problem id start panicking
oh yeah also 130 and 140
and 170
almost all of them work
the extra factor of gcd doesent do anything, /phi(gcd(20130,k)) should be fine as it actually doesent increase the prime factors of phi(2013m), the only worrying thing is phi(k) but it seems like 1<=k<=30 the only troublemaker ks k=29 as phi(29)=28=7×4 and 2013×10×29 isnt divisible by 7, so thats prolly (l)
oh nice
okay thank you!!
indeed
without the walktroughs and you id never figure this out in a million years lol
ty 
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id never figure this out without the walkthroughs either lol
youre welcome
in case you ever want my help with contest math in the future, just ping
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so
it used
since the sets are distinct (excluding the 1)
isnt it just 1 + number of primes less than 50?
How is a union of a set just a number?
n()
What does n() mean
Still flawed then
@tough vigil is n() defined somewhere
Oh wait
Yep
Union of {1,2}; {1,3}, {1,5}...
so prime less than 50 are 1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47
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can someone help with this
not able to understand the different cases i have to take
like do i have to consider 7+2 and 7+1+1 or just 7+2
So, till (1+x^9) we can get x^9 after that, we cant so the sequence important to us is (1 + x)(1+x^2)... (1+x^9)
yeah got that
after we just have to form cases
1+1 isnt possible as there is only one (1+x)
oh.
so one case is 7+2
no worries have a great day
btw are u a jee aspirant by any chance
this year
oh atb man 🙏
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!help
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!showyourwork
Show your work, and if possible, explain where you are stuck.
The arrow just means that the relation goes from integer to integer
$R = \left{(x, y)| y = 2 x + 1, x \in \bZ, y \in \bZ\right}$
@autumn girder
It's the same as this if thus makes more sense
Have you studied relations @tough vigil
?
it is just !show btw
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find n
!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
this is definitely some jee brainrot
2
!showyourowkr
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😂
i just wrote all the fractions as
1-1/2, 1-1/4,1-1/8
and then tried multiplying -1 inside
and got a total of nothing
@autumn girder could u help me w what to do
Have you tried using the identity $(1 - x)^n = \sum_{i = 1}^n(-1)^i{n \choose i}x^i$?
@autumn girder
Well, you can use the distributive property of multiplication over addition to take the added terms out
and then apply this
huh wait i didnt understand this
how do i take the added terms out
$a(b + c) = ab + ac$
@autumn girder
wait so u want me to distribute that over 5 terms?
dude..
which you can then use this on
oh no, 5 terms which look basically the same
I mean doing that you get $\left(\frac 1 2\right)^n + \left(\frac 1 4\right)^n + \left(\frac 1 8\right)^n + \left(\frac 1 {16}\right)^n + \left(\frac 1 {32}\right)^n$
@autumn girder
Well try completing the problem first
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np
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how do we know that ${\bar{A} \neq \emptyset}$?
k
Who says it is
we don't
it exists because it's given a clear definition
it is the set of all points x ∈ A which satisfy x ∉ g(x)
so g(x) is some function that maps each element of A to p(A), returning a set. we define A bar as a set containing the elements of A not found in any set of the range of g(x). am i understanding this correctly
and A bar must exist because it is clearly defined
?
so we use A bar as a counterexample that g(x) is not surjective
is this a non-constructive proof, im new to proof as well as this
😭
we define A bar as a set containing the elements of A not found in any set of the range of g(x). am i understanding this correctly
no
you're not understanding it correctly
if you want though i can explain the same idea in more hands-on terms
would you like that? @north roost
yes plz
ok rigth
right*
imagine A is a set of people
$\wp(A)$ is then the set of all possible group photos that can be made with members of $A$, including one empty photo'' i.e. $\rien$ as well as portraits'' i.e. singleton subsets of $A$
Ann
mhm
now $g: A \to \wp(A)$ is an assignment of people to group photos. in other words, you can think about it like \textbf{giving} each person one group photo from the big pile that is $\wp(A)$.
Ann
our goal will be to show that we couldn't possibly have given out the entire pile no matter how we do it.
(which is the translation of "g is not surjective" into this setup)
@north roost with me so far?
yes
ok right
now
call a person "happy" if they can see themselves in the group photo they got, and "unhappy" otherwise.
i.e. $x$ is happy iff $x \in g(x)$.
Ann
then $\bar{A}$ is the group photo with all the unhappy people.
Ann
makes sense?
yes
now the thing is
nobody could have possibly received that group photo
if the recipient of this photo is happy they would see themselves in it -- but then, because A-bar depicts precisely all of the unhappy people, it'd mean the recipient is also unhappy.
but if the recipient were unhappy, they wouldn't see themselves in this photo -- and yet they would appear there, because of their unhappiness
so assuming that the "unhappy bunch" photo was given out to anybody at all leads to a paradox
im kinda confused with the conclusion
"but if the recipient were unhappy, they wouldn't see themselves in this photo -- and yet they would appear there, because of their unhappiness"
why would they appear
there if they are not
there
they'd appear there by virtue of being unhappy
the unhappy-bunch photo by definition shows all the unhappy people
but that photo is not given to anyone, right?
with all the unhappy people
let me rephrase it. so the unhappy recipient will have the picture that doesnt have themself in there, but they will still appear in the group picture of unhappy people that is not given to anyone
because if it belongs to another unhappy person, that person is happy
so
there is a group photo of unhappy people somewhere, but it is not given to anyone
right?
😭
@pseudo basin 🙏
that's the point.
see here again
,
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Why is the angle between two vectors equal to the angle between the corresponding radius vectors, if the vectors are perpendicular to the radius vectors?
or this picture, if it's clearer
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Hi.
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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
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erm actually they just said hi
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do you have a question?
@mental pebble Has your question been resolved?
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hi im going crazy
so i have a like-sided triangle with side 3
and i was wondering what its height was
oh wait
i just figured it out lmao
nevermind im not feeling crazy anymore
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Hello
where are you stuck at?
okay umm first we'll factor out the -2 so that the coefficient of x^3 becomes 1
making it -2(x^3-5x^2+8x-4)
ok what happens after that
well there are 2 ways you can do this, either find one root by trial-error or factorise
and by looking at this i could see that x=1 would satisfy the equation (this is the trial-error method where you put values like 0,1, etc and find a root)
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how prove that log(x^x) > x for x >= 10
yea ln
Wait is it true that ln(x^y) = y ln(x) for x >= 10?
ln(a^b) = b ln(a)
I feel like this might follow from calculus , but i cant prove it yet
do they have to be integers or can they be real numbers for sufficiently large positive x,y?
]0;+infty[
Any real positive number
Is it french?
I thought all people using it
alr i think i get it
so once ln(x) becomes > 1 and increasing then ln(x^x) = xln(x) > x
thanks
.close
x,y inside R*+
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i have a few questions about this proof
how do we get that M_j=j^2/n and m_j=(j-1)^2/n ?
Because f is strictly increasing
and M_j is defined to be sup of f on the interval, and m_j is defined to be the inf
so the sup would be f of the rightmost point, and the inf would be f of the leftmost point
it should be (j/n)^2 = j^2/n^2 really
but isnt the left and rightmost points 0 and 1?
Of each interval created by the partition I mean, not [0,1]
hm
im still not seeing how they came up with those specific values
each subinterval is length of 1/n right?
the partition is {j/n} for 0 <= j <= n, so each subinterval is [(j-1)/n, j/n]

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Guys any ideas on this
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PA
Point of inflection and rate of change
Found by defining the graph
Graphing it
And analyzing
With find inflection
To find rate of change at a certain point
Dy/dx(f(x)) | x= point value
What you need to do
Understand how to graph peace wise functions and start the second gragh at the point where the drug is no longer effective.
Common words for the analysis in worded form
And info can be in dot points
PB
Define equation
Graph equation
Menu analysis
Inflection
Is the maximum value higher than the overdose cap yes or no
No
Graph additional equation from point of inflection
And if the next maximum isn't greater that the overdose cap continue.
Yes
Once the next pill passes the overdose threshold delete that graph and start next graph at the point the drug is no longer effective in the bloodstream
Repeat until you reach the specified domain. Of the question
Count the maximum amount of pills allowed to be taken within the domain(for eg 24hs or 12h)
What you need to learn
How to connect and graph the equation in a piece wise function with 5 to 10 graphs
Fastest ways to write them out
Using the define functions
In order to connect each together
Understand how to explain what you've done and the questions in relation to it.
This is the information I know about it .
I need help Graphing drugs on Cas in relation to log and exponential.
With something like this
f(t)=A×e^-bt ×In (bt+2)
And. I was wondering if anyone had some practice questions like it. Or information that could help me
@deft token Has your question been resolved?
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<@&286206848099549185>
The single question is at the bottom
I need help Graphing piece wise functions on Cas
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help, looking to confirm my answers
@wise nacelle Has your question been resolved?
,calc 500 * e^(-0.032 * 3)
Result:
454.23200803435
,calc 500 * e^(-0.032 * 8)
Result:
387.07098439612
,calc 500 * e^(-0.032 * 20)
Result:
263.64621202152
half-life is correct though
no, 454, 387, and 264
yeah it is 264
oh ok tysm
np!
.close
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Help
Is it correct?
yeah. you couldve saved some hassle imo if you just completed the squares:
x^2 + 10x + 25 + y^2 - 6y + 9 - 25 = 0
(x+5)^2 + (y-3)^2 = 5^2
centre is (-5, 3)
radius is 5
but thats completely personal preference
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oh lol 😭
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Yes
$\cup$
Yay
Carbonite
its this btw
Light
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lock in bro
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I learned this just few days ago and im struggling with solving by myself, can I get a guide just for the 1st one
hmm what do u mean
i know i have to use P to find max/ min rate of change
I dont exactly know what to do with 2,1 vector
you have to normalise that vector
normalise your vector, find the directional derivative and see what result u get
on the basis of that u should get ur answer
what does normalizing the vector mean, cant i use it for directional derivative equation directly?
normalising the vector means converting your vector into a unit vector
the derivative equation requires u to find the unit vector because converting it into a unit vector limits its length such that the speed of change only depends upon the direction of the vector and not how long it is
if u get what i mean
that makes sense
yup yup
so I find the direction of 2,1 vector, i find the change of direction of the equation? and compare
like max, min rate of changes
sorry i soned out
lol np
yes
yeah, follow it by normalising direction vec and the rest of the steps
u should be fine
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what are integers
whole numbers, positive, negative or 0
..., -2, -1, 0, 1, 2, ....
or do you mean more philosophically?
integers
1
these
oh
anything without a decimal
So whole numbers and integers same thing?
sometimes people say whole numbers to mean just positive (and maytbe 0)
its not really a technical term
So no fractions and decimals = whole number...
yes
Wait huh
dont stress it tbh
But do integers include fractions
no
nope
2/2 
real question
whole numbers are nonnegative integers
how did u answer this
but dont know what an integer is
listen
you have those mixed around
BUDDY IS COOKED
Eh?
wait that is a good question how do u know what a directional derivative is and not an integer we havent even learned this in class 😭
to answer it simply
I studied ahead because I wanted to
😭🙏
he so deep in the rabbit hole he forgot the entry
to answer it simply, bro used mathcord instead of google
REAL
and u called me the nerd thats crazy
so im working with 3d geometry
😔
and i dont know what ts called but imma write it out
What
vauge description and don't include negatives
ah damn whole numbers are positives?
Natural = 1,2,3,...
Whole = 0, 1, 2, ....
Integers = ..., -1, 0, 1, ...
shi mb translation error
O
What about rational then
yep what he said
We love mathcord 🔥
any number that can be represented as a/b for integer a and b
Technically that's any number
basically combining integers and fractions
you cant write sqrt(2) as a/b
sqrt2 isnt an integer
How not
whole numbers are integers
guys how do i go from a Parametric equation to a Cartesian equation of the line
is that right had to ask deepseek for translation?
sqrt2 is not an integer like skissue said
how not
what's the problem with it
chill i thought yo question answered
cuz.... its not rational??
ok how about this
what
rational numbers are fractions
and check if its rational
I might as well do my every question on calculators and not bother solving them 😭
yes
light maybe instead of studying ahead of the class and looking at directional derivatives which arent even gonna be in our final u should touch up on basic set knowledge 😭
I'm doing it today man </3
light how old r u
nineteen
dayumn
noo light learn all directional derivatives and solve my questions ☺️
proof by contradiction
I did it in grade 10
integers do not have any fractional or decimal part to them
YEAH
so fractions aren't integers?
no
well
2/2 is an integer
because it is 1
2/3 is not an integer
because it has a decimal part to it
so fractions where a/b are different and b isn't a factor or multiple of a?
no
oh right
Technically every number as a decimal point
2.00 is still an integer
if you are a cs student, u would call this a double
but whatever
mia u do cs?
no
oh ok
I see 🤔
hes just ragebaiting atp
and furthermore
no
I'm not
Mathematics takes understanding
Sybau
And and
common this is like standard math level
mods ban him
Rational numbers are real
jk
yes?
yes
so are integers?
are irrational real?
yes
yes
alr so lemme make a graph of this
we reinventing xy axis with this one🔥
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noo its not interesdting 😭 icl pmo 🥀
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AD = AB
What’s length for AC?
I am stuck at this question idk how to start
<@&286206848099549185>
do you know cosine rule?
Yes
then you can use that in the triangle ADC
How
You know the angle ADC = ADB+BDC
Yes
you also know CD and can calculate AD
How
pythagoras' theorem???
Okay
What do I do after finding AD and ADC
Oh
Cosine therome
So many steps
Is there easier way
Thanks thou
idk cant think of one on the fly rn sry
cos rule is the first think that came to mind seeing this
So pro
Okay thank you so much for helping me
np
Have fun on the fly
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hey
how would I do b
I can't use the discriminant
I did dy/dx and got 4x^3 + 6x^2
then set it to 10
turned that into 4x^3 + 6x^2 - 10 = 0
but got 2 solutions
,w D[x^3 * (x+2), x]
,w graph 2x^3(2x+3)-10

sounds like the question is wrong
i dont think so lol
wait
this is the correct working
,w graph 4x^3 + 6x^2 - 10
does this look correct
3x^2(x+2) + x^3 = 3x^3+6x^2+x^3 = 4x^3+6x^2
This graph is incorrect
You wrote 2x^3(2x+3) - 10 but its 2x^2(2x + 3) - 10
i still dont get why i have 2 solutions
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I would like it to be as general as possible. Say, I got a matrix and a function.
How should I compute the resulting function after applying the matrix?
Let's say I got y=1/x, and the matrix to be:
wdym by applying the matrix?
the expression 1/X is not well-defined for a matrix X
You need to define it yourself if you want to give it some sort of sense
Like considering the matrix as a transformation. I wonder what should I exercise to obtain the function after transformation.
Let's say x belongs to real number but not 0
I just wonder what is the common method to gain the function after transformation.
Confusion
Also, I would like to learn the rationale behind the method as well.
how do you transform a function by a matrix lmao
Function is a graph. Transformation alters the graph.
From what I'm getting, you seek to obtain a set of transformed points after applying that matrix on the points (x,y) lying on the graph of y=1/x, correct?
wait nvm i get it now
Exactly, sir.
Just multiply the matrix M with the column vector (x,y)
And put y=1/x
You'll get the set of transformed points, but not a function per se
I will get a transformed column vector of (x,y) after multiplying it by matrix M.
Yeah
By putting the transformed vector, let's say, (x',y') into y=1/x. I should get the transformed function.
Computing 1/v where v is a vector is not well defined in general
And the latter part of your argument doesn't make sense
By the latter part, you mean the result will not be a function?
There will be no result if you put vector v = (x',y') in f(x) = 1/x unless you resort to exclusive definitions for 1/v
What is your original complete question?
I will send it immediately, sir. Give me one second
I WILL TRANSLATE THEM. IT IS A QUESTION FROM COMPETITIVE MATH.
No it's not translation
It's this
And you can see after applying the matrix, you get a point (and to be exact, a set of points for non-zero Real x)
set of points, as a whole , should be a function.
A set is not a function
A function is a mapping or relationship between sets (with more conditions)
May I know what elicits the qualification?
What you are doing here is applying M on every point (x,y) lying on y=1/x
You apply a matrix on a point (specifically, the position vector)
You're doing this
$$\begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$
à뜜
The matrix should convert the graph. The resulting (x',y') should be considered the graph after the transformation. That is, (x',y') display the transformed function on the xy plane.
@helper
我看得懂:),so you mean MN(x, y) where y=1/x
<@&286206848099549185>
that is acting on R2 and adding a constraint
you consider the graph of 1/x as a subset of the plane is that it?
Glad you understand it, and yes, that’s exactly what I mean
Matrices got remove in my syllabus this year
yes or algebraic variety as you may
you could apply the matrice to this vector
so what theyre saying is that you rotate the whole plane, and then you draw out the rotated curve
Exactly
Since it’s a rotation matrix in this case
this is correct
but then figuring out the resulting curve could be tricky
the point is that imagine (x', y') is the rotated set of points(curve), so every single original point (x, y) obeys y=1/x, and you rotate each point to the resulting point
One way is to express (x - 1/x , x + 1/x) in the form (t,g(t)) if we are finding the closed form of rotated graph
is that the justification youre asking?
Yes. Thank you sir
this rotate
yep and stretch a bit
the general solution is like this, in the video you posted the solution assumes the form x'^2+ay'^2=b since we know y=1/x is hyperbola
the answer follows since b does not depend on x' or y', so a is set to be -1
wait nvm
there is, by factor of sqrt(2)
Yeah
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Linear approximation
Should be cos(0) in the gradient
What is this supposed to represent
linear approximation?
y^2 is not linear
Show the full page where you got this from
here's whole exercise
there's nothing but this line
Then there is only a partial answer
This is really two questions
I already solved (a)
Only (b) left.
Everything else on these pages is just other exercises.
Yea so like I said above, they only provided a partial answer
Use Taylor Theorem?
Development of Taylor's polynomial for functions of many variables.
@unique vortex Has your question been resolved?
Seems like you're not wrong, it's just the second order translation messing with you
*grad f dot (x-a)
@unique vortex
Ty, got the same answer. Messed it up due to two dot products once though.
grad(x,y,z) really helped with hessian matrix
How to calculate the Taylor polynomial
of degree two?
The nth term of a Taylor polynomial is $\frac{f^{(n)}(z)}{n!}(x-z)^n$ where z is where you're centered at
Parsec
So, (1,0,0) is center?
Oh nevermind that's only for single variable
essentially the same thing
So, linear approximation and second degree tp are the same thing?
no i mean multivariate takes similar form as single variate
i think you're looking for frechet derivative
i saw that before but not familiar sorry
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yo guys

