Can you use that here?
#help-42
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verbal abyss
mhm, we have the mean of 70 and the sd of 10
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So what are the 3 intervals A is asking for?
verbal abyss
I dont think im supposed to be calculating anything tho, the questions given dont seem that way to me at least
o
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I dont think im supposed to be calculating anything tho, the questions given don...
Can you find the 7 numbers near the diagram in your notes?
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Yeah, for this problem
calm coralBOT
verbal abyss
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verbal abyss
wait so
since carla got an 82 on the exam, its a little off of the class avg (mean) of 70. So hers score would be interpreted as slightly above average?
Im still lost on what question A is asking D:
nimble badge
Yeah that’s good
nimble badge
Im still lost on what question A is asking D:
Question A is a bit more hypothetical, what does it mean to be within 10 points of the mean? 20? More than 20?
verbal abyss
Mean would be what most ppl got so 10 more/less would mean someone got a score higher/lower than 70
nimble badge
Mean would be what most ppl got so 10 more/less would mean someone got a score h...
Well most people got within 10 points on either side of 70
verbal abyss
I dont understand v__v
The answers are right in front of me but im not connecting them at all 😭
nimble badge
So you got B, I think you can get C, A is a theoretical question of how to interpret a probability distribution generally
nimble badge
The answers are right in front of me but im not connecting them at all 😭
So maybe this would help: suppose only Carla took the exam, the mean would be 82, if Kyle takes the exam the mean would be (82+66)/2 if the next person takes it (82+66+94+…)/n if you took many samples (infinitely many) and plotted a histogram of their exam scores it would look like that bell shape. We want to use numbers to describe this bell shape, so we use standard deviation to measure how far away from the average result an individual student’s score is
And when the distribution is a particular way, we can use Z-scores to make the standard deviations easier to work with (in the same way that 82/100 is easier read as 82%)
verbal abyss
mmmmm okokok I think i have a form of understanding now, I appreciate your help a ton ^-^ I think i can do c as well on my own
have a great day/night :)
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sharp patio
Did I solve the problem correctly?
Find the work done by the force field F(x,y) = <x^2,-xy> in a moving particle along r(t) = <cos(t),sin(t)>, 0<=t<=pi/2
calm coralBOT
Did I solve the problem correctly?
Find the work done by the force field F(x,y)...
@sharp patio Has your question been resolved?
sharp patio
<@&286206848099549185> would appreciate any help as someone else got a different solution but I'm unsure where I went wrong :p
I did it another way and got -2/3, however I'm not allowed to do it that way 
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faint dagger
How to do
limpid pilot
How to do
write a1 and a20 as a + (n-1)d
you'll get 2 equations
use them to find d
then just find S21
faint dagger
Ai lemme try
night dragon
,rotate
potent lotusBOT
twilit bramble
calm coralBOT
twilit bramble
Need help with g and h
hasty fiber
,rccw
potent lotusBOT
night dragon
for g
consider that 3 and 5 have to be included
so you only have to count their different permutations within the number
and then you have 3 choices for the last number
twilit bramble
and then you have 3 choices for the last number
Ok so would it be 3 x 3 x 3 over 5P3
night dragon
Ok so would it be 3 x 3 x 3 over 5P3
not quite
think of it this way
5 and 3 have to choose their places
so it would be 3c2 * 3/5 * 4 * 3
calm coralBOT
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night dragon
What abt h
consider the separate cases where it only contains 3 and only contains 5 and add them ogether
the question is ambiguous
is it only 3 or 5, or does 3 and 5 also work for h
calm coralBOT
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twilit bramble
is it only 3 or 5, or does 3 and 5 also work for h
I think it’s just 3 or 5
Not both
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jade snow
probably a stupid mistake but what did i do wrong
calm coralBOT
probably a stupid mistake but what did i do wrong
jade snow
isn't the magnitude formula just square root the sum of all the terms squared?
nvm i figured it out 💀 didn't need to derive v(t) before squaring it
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sour frigate
can i find the maximum of a cubic function without calc
for a quadratic its just -b/2a but ig its not the same for a cubic function
glass heart
a cubic function has no (global) max
and its not guaranteed to have a local max either
glass heart
then the max might often be at the ends of the interval
sour frigate
isnt there a formula or something
glass heart
well you can find one by differentiating and then solving the quadratic
and then you can forget that you used calc
and now you have a formula that you can use without calc
sour frigate
lmfao
im in precalc rn
i have a test friday and im wondering if my teacher is mad enough to ask us to find the max on a cubic
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meager verge
find the image of the point (2,-1) under the sequence of transformations: a translation of 3 units in the positive direction of the x-axis followed by a dilation of factor 2 from the axis
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find the image of the point (2,-1) under the sequence of transformations: a tran...
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noble rock
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noble rock
i have used combinatorial identity but that didnt solved this
answer is sqrt 3
plz help anyone
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noble rock
<@&286206848099549185>
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noble rock
<@&286206848099549185> im unbale to to solve can u pl zhelp
dusky fjord
Yes
jaunty hawk
hi
guys
im a profeesor in IIT Bombay, India
im 59yrs old
my body cant wokr
work
i have to speak while standing
@everyone
noble rock
sry?
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buoyant mirage
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mortal orbit
full q but i can do lhs not rhd
$\sum_{r=n}^{2n}... = \sum_{r=1}^{2n}... - \sum_{r=1}^{n-1}...$ if that's what's bothering you
potent lotusBOT
rafilou2003
mortal orbit
full q but i can do lhs not rhd
how did you do lhs then?
mortal orbit
the lhs is def wrong
buoyant mirage
yeah 😭
mortal orbit
do you know $\sum_{k=1}^n k$, $\sum_{k=1}^n k^2$, $\sum_{k=1}^n k^3$?
potent lotusBOT
rafilou2003
mortal orbit
that's 3 sums
I was wondering if you knew what each one was equal to
$\sum_{k=1}^n k =$ ?
potent lotusBOT
rafilou2003
mortal orbit
ok so you DO know it
mortal orbit
tahts cuz i memorized it idk the logic behind it
Name S = 1 + 2 + 3 + ... + n
then S is also equal to n + (n-1) + ... + 3 + 2 + 1
buoyant mirage
ohh r u talking abt arithmetic total rule
mortal orbit
$S = n + (n-1) + (n-2) + ... + 3 + 2 + 1\
S = 1 + 2 + 3 + ... + (n-2) + (n-1) + n$
potent lotusBOT
rafilou2003
mortal orbit
so $S + S = (n+1) + (n+1) + .... + (n+1)$
potent lotusBOT
rafilou2003
mortal orbit
so $2S = n(n+1)$
potent lotusBOT
rafilou2003
mortal orbit
so $S = \frac{n(n+1)}{2}$
potent lotusBOT
rafilou2003
mortal orbit
tahts cuz i memorized it idk the logic behind it
so that's the logic behind it
mortal orbit
ok fine this helps w rhs
no the RHS is not this, the RHS uses the sum of k^2
maybe you know what $\sum_{k=1}^n k^2 =$ ?
potent lotusBOT
rafilou2003
mortal orbit
no
buoyant mirage
then
mortal orbit
$\sum_{k=1}^n k^2 =\frac{n(n+1)(2n+1)}{6}$
potent lotusBOT
rafilou2003
mortal orbit
so if my n was 2n i would sub 2n on rhs
not exactly, as I said the sum in the RHS goes from r= n to 2n
it doesn't start at r = 1
buoyant mirage
ur so right
mortal orbit
**rafilou2003**
this is where my first message comes in
buoyant mirage
n-1((n-1)+)(2(n-1)+1)/6
mortal orbit
$\sum_{r=n}^{2n}r^2 = \sum_{r=1}^{2n}r^2 - \sum_{r=1}^{n-1}r^2 = \frac{(2n)(2n+1)(2(2n)+1)}{6} - \frac{(n-1)((n-1)+1)(2(n-1)+1)}{6}$
potent lotusBOT
rafilou2003
mortal orbit
final help you should need : $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$
potent lotusBOT
rafilou2003
mortal orbit
and now you should be able, thanks to the 3 formulas I gave you, to find the LHS
mortal orbit
how can you relate a sum of (r-1)r(r+1) to a sum of r^3, r^2 and or r?
mortal orbit
yep
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potent raven
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potent raven
How do I show that this function is continuous în x0=3
The funnction is defined as such
f:[0,3]->R
I understand that I have to do the limit from the left = the limit from the right = f(x0)
But how do I find the limit from the right if the rightmost point of the function is already covered?
Is it that that point itself is the limit?
<@&286206848099549185>
untold compass
But how do I find the limit from the right if the rightmost point of the functio...
yea at the endpoints you just need to show one side of the limit is the point itself
since the other side just isn't there
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last zealot
I need help with a physics problem
calm coralBOT
I need help with a physics problem
last zealot
I'm trying to graph planck's law of radiation derivation
But my graph is just completely flat when I write it out
this is what it's supposted to look like
Why isnt it working
Ive sat here for 2 hours
I can't get it to work
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worn sandal
Anyone know how to simplify this
calm coralBOT
Anyone know how to simplify this
worn sandal
We are given (X_i, Y_i) are iid
And that X_i = (X_{i,1}, ..., X_{i,K}) where each X_{i,k} = 1{X_i = k}
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untold compass
Anyone know how to simplify this
if X and Y are independent you can remove the | Xi=k part, since it doesn't influence Y at all
worn sandal
if X and Y are independent you can remove the | Xi=k part, since it doesn't infl...
X ad Y are not independent
untold compass
oh woops
worn sandal
we have the regression Y_i = \beta_1 X_{i,1} + ... + \beta_K X_{i,K} + U_i
<@&286206848099549185>
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wintry copper
The question says: For each part: Graph the parent function, and each transformed function. Show rough work at the bottom or on extra paper. Describe the effects of each variable on the graph of the function. Think in terms of the following features, as well as any observations that stand out
wintry copper
ivory saddle
What's the problem? Do you know what is meant by a and k?
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wintry copper
What's the problem? Do you know what is meant by a and k?
Nah
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remote mural
i just wanna make sure of something, but for doing curls using the quick-hand determinant method of [
\curl \vj A = \mat[v]{\vcx &\vcy &\vc*z \ \partial_x &\partial_y &\partial_z \ A_x & A_y & A_z}
]
are we implicitly expected to always take the cofactor expansion across the first row? is it possible to take the cofactor expansion across any arbitrary column or row?
potent lotusBOT
remote mural
I believe the main thing i heard in relation to this is that since the determinant formula for cross products is reaaally not a true determinant and rather some quick-hand way of remembering how to calculate cross products, taking the cofactor expansion willy nilly doesnt really work
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oblique current
@remote mural yeah this is rly just a shorthand way for doing the calculation
There will be some diff geom reason why it resembles a determinant probably but it escapes me
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knotty panther
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knotty panther
what happened here
eternal shard
there is this sum rule and they factored x^n
vivid vine
you can combine sums like that, and x^n got factored after
knotty panther
the have to have a common variable to get added ?
knotty panther
there is this sum rule and they factored x^n
what's the sum rule
eternal shard
basically combining the sum terms into one
but they need to have the same indexing
knotty panther
but then that way i cant take x^n as a common factor here
vivid vine
if you mean a variable as 'n', they need to be the same, and to have same bounds here for example n=0 to inf
the x^n is still inside the sum
potent lotusBOT
𝔸dωn𝓲²s
knotty panther
i dont really get what you're saying here
how did we take x^n as a common factor when x^n wasnt common in both terms
vivid vine
combine the sums and tell me how the sum looks like
before any factoring
beacuse x^n is definitely a factor in both sums
vivid vine
np bro
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cedar ether
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cedar ether
Apparently the "only if" part is very easy and left to the reader. I struggle with this; in other words, if $f$ is measurable, why are $f_1$ and $f_2$ measurable?
potent lotusBOT
Philip
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remote mural
Apparently the "only if" part is very easy and left to the reader. I struggle wi...
Are you asking why the Cartesian product is measurable or the other way around?
Well I'd assume because it checks the same conditions for being a sigma algebra.
Say if X and Y such that (X,Y) is an element of f.
You want to check if the complementary (X,Y)⁰ belongs to f.
And if the union of all (X,Y) belongs to f. If that's the case you can figure that separating X being an element of f1 and Y element of f2.
That will lead you to conclude F1 and F2 are sigma algebras too.
Thus measurable sets.
cedar ether
Are you asking why the Cartesian product is measurable or the other way around?
what I'm trying to figure out is if $$\forall B_1\times B_2\in \mathcal B_1\otimes\mathcal B_2, \ f^{-1}(B_1\times B_2)\in\mathcal A,$$ does it then hold that $$\forall B_1\in \mathcal B_1, \ f_1^{-1}(B_1)\in\mathcal A\quad\text{and}\quad \forall B_2\in \mathcal B_2, \ f_2^{-1}(B_2)\in\mathcal A?$$
potent lotusBOT
Philip
remote mural
Yeah
proving the other way of the equivalence
Like I said, try using the definition of sigma algebras
This time for the set B1 x B2 of F
For example, what is the complementary of (X,Y)?
Is it the Cartesian product of the complementary of both X and Y?
cedar ether
I'm thinking, we have $$f^{-1}\left(B_1\times B_2\right)=f_1^{-1}\left(B_1\right)\cap f_2^{-1}\left(B_2\right).$$ If $f$ is measurable, then this set is in $\mathcal A$, but does that imply that $f_1^{-1}\left(B_1\right)$ and $f_2^{-1}\left(B_2\right)$ individually are also in $\mathcal A$? All we know is that their intersection is in $\mathcal A$.
potent lotusBOT
Philip
remote mural
I don't think the cross product of B1 and B2 is the intersection
I see what you mean
$f^{-1} (X\times Y) = f^{-1} (X) \times f^{-1} (Y)$
potent lotusBOT
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cedar ether
$x \in f^{-1}(B_1 \times B_2) \iff f(x) \in B_1 \times B_2 \iff f_1(x) \in B_1 \land f_2(x) \in B_2$
potent lotusBOT
Philip
cedar ether
so $x \in f_1^{-1}(B_1) \cap f_2^{-1}(B_2)$
potent lotusBOT
Philip
remote mural
You still have to prove that F1 and F2 are measurable.
By the definition of measurable sets. What you're doing is proving the element x of (B1,B2) is in B1 and B2
Which holds true but it's not enough for F1 and F2 to be measurable sets.
Is the complementary there?
And the union of all x too?
cedar ether
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knotty panther
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knotty panther
what is this identity ?
potent lotusBOT
y0shi
glacial inlet
and they just rearranged and solved for sin^2(x)
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royal raft
How did they get infinify when doing l'hopitals rule? I got 3 which is still diverges
royal raft
if you expand (n+1)^2
you get n^2 + 2n + 1 right
and you have n^3 over that
oh
I see where I went wrong
I took a derivative when I didnt have an indertermine form
I got 3n now
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onyx quiver
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onyx quiver
how do i draw f'(x)?
untold compass
find the 0 points and try to guess the slope of the highest inclination point, then just draw another curve like f(x) waving up and down through them
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lean bluff
I'm asked to find the antiderivative of ∫(3x-1)^2dx. Is it permissible to distribute the (3x-1)^2 into (9x^2) - (6x) + (1), and then to find the antiderivative of each of thse in accordance with the sum rule, then combine them?
charred coyote
Yes integral is linear so it is allowed to distribute these
You can also use the method of u substitution
lean bluff
I'm not too familiar with u substitution? How could I use it in this situation? Thanks, by the way!
charred coyote
You can set u = 3x-1
Then you get du = 3dx
You can then evaluate the integral as 1/3 integral(u^2 du)
Then you can variable sub x back in
This is motivated because we know the integral of x^2
U-sub is an alternative name for change of variable since often the variable chosen is u
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lean bluff
Actually, another question. If I do use u substitution, I come out with 3x^3 -3x^2 +x -1/9 but I'm entirely unsure where I'm getting the 1/9. I've solved this both using u substitution as well as distributing and combining, but I end up with simply 3x^3 -3x^2 +x when I distribute.
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full epoch
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full epoch
Would this just be like y=a(3)^1/9?
wanton wave
Would this just be like y=a(3)^1/9?
close, but aren't you missing an x somewhere?
full epoch
Uhhhh i have no idea
wanton wave
You've got the right idea, why 3^(1/9)?
full epoch
Triples by 3 every 9 years so multiply by the power of 1/9 to cut it down idk lol
wanton wave
Yeah, that's right
If it multiplies by 3 every nine years, then it multiplies by 3^(1/9) every year
wanton wave
so what's the yearly growth factor?
full epoch
1.13
full epoch
So then is it a 13% change each year
wanton wave
yep 👍
full epoch
Ah ok cool thanks
wanton wave
np 👍
full epoch
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calm coralBOT
wanton wave
Would this just be like y=a(3)^1/9?
the actual equation would be y=a(3)^(x/9) though
full epoch
Well yea ik i just mistyped
wanton wave
or $y = a(3^\frac{1}{9})^x$
potent lotusBOT
tatpoj
wanton wave
but yeah the growth factor is b=3^(1/9)
full epoch
Then whats the value of x
wanton wave
x is any number of years, and y will be the number of fish in the pond after that many years pass
The question didn't ask you for any of that though
I was just clarifying
full epoch
Oh ok
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zenith mirage
I was trying to understand the proof of Kruskal algorithm. I am not able to understand what it means by "that agrees with T for a longer initial list of edges than T* does".
zenith mirage
Is it trying to show the fact that there exists another T1 that has more common edges with T than T* has and the weight(T1) <= weight(T*)?
So, there will be a point when weight(T)<=weight(T1)<=weight(T2)<=...weight(Tstar)?
And since we said that T* is an MST therefore, all of these should be equalities? So, Kruskal's T is also an MST?
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quiet rover
i need help in solving this inverse laplace transform : find L^-1(1/(s-1)^5)
quiet rover
<@&286206848099549185>
exotic falcon
<@&286206848099549185>
!15mins bruh
calm coralBOT
<@&286206848099549185>
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
exotic falcon
What have you tried
quiet rover
i tried convolution thm
is the ans e^5t ?
<@&286206848099549185>
<@&286206848099549185>
hello ?
just tell me am i right or wrong
quaint sphinx
is the ans e^5t ?
the convolution of e^t and e^t is not e^2t
quiet rover
i dont understand
quaint sphinx
i assume when u applied convolution theorem u had something like e^t * e^t * e^t * e^t * e^t
my point is this expression does not equal e^5t
quiet rover
i assume when u applied convolution theorem u had something like e^t * e^t * e^t...
yea
quiet rover
my point is this expression does not equal e^5t
ok
quaint sphinx
the * means convolution, not multiplication
(though to be honest i would not be doing this problem with convolution theorem anyway unless your teacher is asking you specifically to do so)
quiet rover
so i have to apply convolution ?
quaint sphinx
if u want to do this problem using convolution theorem, then yes
quiet rover
(though to be honest i would not be doing this problem with convolution theorem ...
well in the question it just says find and doesnt specify any thing
quaint sphinx
well then don't use convolution theorem
quiet rover
is there an easy way cause its suppose to be a 2 marks question
quaint sphinx
is there an easy way cause its suppose to be a 2 marks question
yes pretty easy
apply s-axis translation theorem
quiet rover
i did F(S)= 1/(S-1)^5 but i dont know what to do further
what formula should i apply here
<@&286206848099549185>
help
me
please
i dont
know how
to solve
this
jolly oyster
im pretty dumb but
would you
shift it 1 unit along the s axis to get to 1/s^5
then use laplace transform formula and get t^4
and because it was transalted 1 put e^t
quiet rover
bro i am dumber
jolly oyster
idk how to write fractions and stuff in this discord
quiet rover
can you send a pic
quiet rover
thanks
jolly oyster
well thats the answrer i ended up with but i really dont know lmao
you could try put it into symbolab and see what it gets
just to check your answers and stuff
quiet rover
i tried chat gtp and it gave me this : Certainly! To find the inverse Laplace transform of ( \frac{1}{(s-1)^5} ), we'll use the formula for the inverse Laplace transform of ( \frac{1}{(s-a)^n} ):
[ \mathcal{L}^{-1}\left{\frac{1}{(s-a)^n}\right} = \frac{t^{n-1}e^{at}}{(n-1)!} ]
Here's how we apply this formula step by step:
-
Identify (a) and (n):
From ( \frac{1}{(s-1)^5} ), we can see that (a = 1) and (n = 5). -
Apply the formula:
Substitute (a) and (n) into the formula:
[ \mathcal{L}^{-1}\left{\frac{1}{(s-1)^5}\right} = \frac{t^{5-1}e^{1t}}{(5-1)!} ] -
Simplify:
Simplify the expression:
[ \mathcal{L}^{-1}\left{\frac{1}{(s-1)^5}\right} = \frac{t^4e^t}{4!} ]Where (4!) is (4 \times 3 \times 2 \times 1 = 24).
-
Final Result:
After simplifying, we have:
[ \mathcal{L}^{-1}\left{\frac{1}{(s-1)^5}\right} = \frac{t^4e^t}{24} ]
This is the inverse Laplace transform of ( \frac{1}{(s-1)^5} ), which is ( \frac{t^4e^t}{24} ).
potent lotusBOT
airbus_380
quiet rover
yes this one
but i cant find the formula chatgtp used
and i find it hard to trust chatgtp for maths
jolly oyster
well chatgpt normally isnt to smart but i mean if it got the same answer as me thats a good sign
maybe ask what formula it used
symbolab said that
quiet rover
well lets see
Apologies for the confusion. In the first question, I utilized the formula for the inverse Laplace transform of ( \frac{1}{(s-a)^n} ), which is given by:
[ \mathcal{L}^{-1}\left{\frac{1}{(s-a)^n}\right} = \frac{t^{n-1}e^{at}}{(n-1)!} ]
where:
- ( \mathcal{L}^{-1} ) denotes the inverse Laplace transform,
- ( a ) is a constant,
- ( n ) is a positive integer,
- ( t ) is the time variable.
In the problem, we had ( \frac{1}{(s-1)^5} ), so ( a = 1 ) and ( n = 5 ). Substituting these values into the formula gives us:
[ \mathcal{L}^{-1}\left{\frac{1}{(s-1)^5}\right} = \frac{t^{5-1}e^{1t}}{(5-1)!} = \frac{t^4e^t}{4!} ]
After simplification, we get ( \frac{t^4e^t}{24} ) as the result.
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airbus_380
quiet rover
it says this
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idle fractal
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idle fractal
Am I right or wrong?
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remote mural
seems wrong to me\
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quiet rover
i need help in solving this
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someone
please
for the love of god
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blazing coyote
Let $|z_1|=12$ and $|z_2-3-4i|=5$ find the miniumum value of $|z_1-z_2|$
potent lotusBOT
ƒ(Why am. I here)=I don't know
blazing coyote
so |z_1| is $x^2+y^2=144$
potent lotusBOT
ƒ(Why am. I here)=I don't know
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blazing coyote
and $z_2$ is (x-3)^2+(y-4)^2=25
potent lotusBOT
ƒ(Why am. I here)=I don't know
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remote mural
circles
potent lotusBOT
ƒ(Why am. I here)=I don't know
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distance
remote mural
well, if you graph these, its very easy
blazing coyote
well, if you graph these, its very easy
hmm
ok
let me try
thanks
hmm
so I find the equation of the line passing via (3,4)
and (0,0)
find the points of intersection with the circle
and then the distance between them
right?
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dry breach
(\log _(a)x*\log _(b)x)/(\log _(a)x + \log _(b)x)
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(\log _(a)x*\log _(b)x)/(\log _(a)x + \log _(b)x)
vocal sigil
@dry breach you can use logarithmic properties
vocal sigil
Log x base a can be written as log x by log a
dry breach
then
vocal sigil
Do it everywhere in the equation
vocal sigil
What do u get
vocal sigil
U dont need to actually keep them as is and simplyfy the denominator
dry breach
how
vocal sigil
LCM
dry breach
loga x b= log (ab) ?
vocal sigil
Log a + log b = log ab
dry breach
woah
vocal sigil
Is it confusing?
dry breach
Is it confusing?
i was multiply it like log x . log x = log x^2
and this too
please bro
vocal sigil
Here's a hint u can write 1 as logx base x or logy base y or logz base z depending on the situation, use this to simplify a b and c and later substitute them
dry breach
Here's a hint u can write 1 as logx base x or logy base y or logz base z dependi...
for each i have to wrtie log x base x then for next one log y base y for next one log z base z ??
or for all log x base x
for thevalue for 1
vocal sigil
for each i have to wrtie log x base x then for next one log y base y for next on...
.
This
vocal sigil
The power 3 is not for the xyz it's for the whole log u can't actually write it like that
dry breach
then 3 x log xyz
vocal sigil
(Logxyz)^3
dry breach
what
yes
yes
nonon
wait
we have logxyz + log xyz + logxyz as per rule we can write 3log xyz
log (xyz)^3
vocal sigil
we have logxyz + log xyz + logxyz as per rule we can write 3log xyz
Could you send where u are solving
vocal sigil
Yh
Logx + logy + logz can be written as logxyz
And it become (logxyz)^3/(logx)(logy)(logz)
vocal sigil
log (xyz)^2 (log x + log y + log z/log xyz)
Actually u can't write logxlogylogz as logxyz
dry breach
we get (logxyz)^2
vocal sigil
Logx + logy + logz can be written as logxyz
.
dry breach
yeah
dry breach
we get (logxyz)^2
next we get
vocal sigil
log (xyz)^2 (log x + log y + log z/log xyz)
There is a slight error in the denominator
vocal sigil
And it become (logxyz)^3/(logx)(logy)(logz)
Simplify this?
dry breach
yes
vocal sigil
U can write logxyz base x and similarly for others
dry breach
ans
dry breach
U can write logxyz base x and similarly for others
.
coral leaf
I need some help, im learning partial fractions and the lesson says that these three are considered proper fractions because the tops are polynomials of lesser degree than the bottoms, im not quite sure what this means and need help understanding it so that i can correctly identify what fractions are proper and inproper in the future
topaz raft
is this x+i
coral leaf
i dont think so, i believe it is 1
dense fog
What don't you understand? You don't know what degree is?
coral leaf
Ive searched up what polynomial means and i dont quite understand it, or at least the third one
coral leaf
i
yh i does, however the rest of the page it quite dirty and has a lot of random dots on the page so i think its a 1
exotic meadow
Could be i
coral leaf
yh it could be
neon scaffold
doesn't matter if it's i anyway, but the degree is the highest exponent of a polynomial
the third one has degree 2 on top and degree 3 on the bottom
dense fog
Ive searched up what polynomial means and i dont quite understand it, or at leas...
What don't you understand? It doesnt help much you just say you don't understand a definition
coral leaf
the third one has degree 2 on top and degree 3 on the bottom
right that makes a lot of sense, so if i had x-2/x+1 would it be inproper because they have the same on the top and bottom?
neon scaffold
yea
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glad sinew
I dont get it, Ker(T)= {0} means its dim is 1?
because 0 vector is still something? its not null
warm warren
The 0 vector in P₂ is the quadratic that sends the entire real line to 0
It looks like this guy, f ∈ P₂, where f(x) = 0x² + 0x¹ + 0x⁰
warm warren
Yes the trivial kernel has dimension 0
warm warren
There is only 1 “0 vector” in each vector space
If your kernel contains that only (and it always contains at least this 0 vector) then we call it trivial
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remote mural
For the orthogonal, curvlinear coordinate systems $u$ and $v$ which have two vectors defined as
\e{align*}{
(A)u &= \vc*u_1A{u_1} + \vcu_2A_{u_2} + \vcu_3A_{u_3} \
(A)v &= \vc*v_1A{v_1} + \vcv_2A_{v_2} + \vcv_3A_{v_3}
}
I want to find a systematic method to do the following two things:
\e{enumerate}{
\ii Recover expressions of the magnitudes of the components of vector $(A)_v$ knowing the magnitudes of vector $(A)_u$, and vice versa
\ii Recover expressions for the unit vectors of $(A)_v$ in terms of the unit vectors of the vector $(A)_u$ and vice versa
}
potent lotusBOT
remote mural
so my friend told me this is really just change of basis, and what i should be doing is something like
convert v to cartesian, then convert that cartesian representation to u
which is like, fine, i can do the step of representing the basis of our original coordinate system in terms of the standard basis
,,, as an example, spherical would yield:
\e{align*}{ \hat r& = (\sin\theta\cos\varphi, \sin\theta\cos\varphi, \cos\theta) \
\hat\theta & = (\cos\theta\cos\varphi, \cos\theta\cos\varphi, -\sin\theta) \
\hat\varphi & = (-\sin\varphi, \cos\varphi, 0)
}
potent lotusBOT
remote mural
after taking the partial derivatives with respect to $r, \theta, \varphi$ respectively of
[ (x, y, z) = (r \sin\theta \cos\varphi, r\sin\theta\cos\varphi, r\cos\theta)
]
potent lotusBOT
remote mural
which is what defines spherical coordinates. So like, cool, we converted our spherical coordinates to be in terms of thr standard basis. Now what? I am a bit confused how to carry on this process to find a representation for the magnitudes of (A)_v in terms of the vector (A)_u, and likewise for the unit vectors
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hard void
How can I do part a, following the method that you integrate y, differentiate x and put the dx in the integral of y
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covert fulcrum
How far have u gotten
hard void
64(sin^2 )t(cos^2)t+48(sin^2)t(cos)t
covert fulcrum
theres a another way of writing sin^2 t and cos^2 t in terms of cos
hint: rearranging double angle
covert fulcrum
after the step i said before, youll get cos^2 2t
somewhere in that
and then u can apply the rearranging again with double angle
youll get cos 4t
covert fulcrum
hmm after changing the intial sin^2 t and the cos one
you should get 16-16cos^2 (2t)
to get this result u should have
64 (1/2 - 1/2 cos2t)(1/2 + 1/2 cos2t)
hard void
yeah i got that bit
my expansion must be wrong
ahh its 1 - cos^2(2x)
ur right
ok so from here how do i finish it
covert fulcrum
ahh its 1 - cos^2(2x)
so it should be 16 mutiplied by that
you can use double angle and reaarange again
16 (1-(1/2 +1/2 cos4t))
then you will get the result :>
hard void
Thanks for ur help
That should have been a lot easier
lol
this is still here so i'll ask smth else
how to integrate
48(sin^2)tcost
nvm reverse chain rule
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onyx quiver
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hasty fiber
that's part of determining whether a point is a local max or min
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pallid mango
oh sorry i didnt get enough time to explain why it's false, the channel was closed 😅
Let's assume it's true that the black angle is 45 degrees. That would mean that the lengths of the legs of the triangle are the same, so the length at the top of the triangle is 4m, just like on the side.
Now, I'm sure you can see why the triangles next to it should also be 45 degree right-triangles. However, if you look at the triangle in the top right corner, the lengths of its legs cannot be equal, which is a contradiction because 45-degree right triangles must have equal legs! It's simple to show that the leg on the very right must have a side length of 6m, but the leg on the top of the triangle must have a side length of 11 - 4 = 7m
therefore, it must be false that the black angle is 45 degrees
limpid pilot
you cant just conclude that by looking at the diagram
unless it explicitly states that
pallid mango
it was given that the outside rectangle and the inside rectangle are both rectangles so they have 90 degree corners, but that was all
pallid mango
you can't, i proved that it's actually impossible for them to be 45 degrees! I think they just assumed it because the diagram makes it look lke 45 degrees, because if the outside rectangle was 10x10m instead of 10x11m then it would actually be 45 degrees
but because the outside rectangle is slightly off from a square it cannot be 45 degrees
clear kelp
hmm
pallid mango
here's the problem for reference
okay but you can't be sure that the angle bisector of the inside rectangle's angle is perpendicular to the edge of the room
in fact, the diagram is totally misleading because the actual angle of the rug looks nothing like the diagram
(if you find the correct answer)
because there's nothing in the problem statement that guarantees that either the top leg of the top left triangle has a length of 4m, or that the carpet's aligned 45 degrees from the edges of the room
i won't give you the angles or side lengths but the solution of the problem does look something like this, if you need some visual intuition why you can't assume that the angle is 45 degrees
obviously the angles of the triangles here are not 45 degrees. and also it's a bit more obvious that the room does not act like a square, cause in the diagram they give you it looks like a perfect square
yeah basically
but nothing in the problem statement and none of the measurements actually say anything that would cause the angle to be 45
what do you mean ?
if you mean that the angles are also not to scale then yeah . they're totally off
yeah
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remote mural
Is the function that returns the second maximum lipschitz continuous
remote mural
Not sure if this satisfies the definition since the maximum change by replacing any coordinate is a constant dependent on other (untouched) coordinates
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limpid mango
can someon eplease help me
i need answer now cuz its due at 2:10
and can you. help me with explanation after
lucid umbra
Sigma notation means a sum starting from i = 1 to n, so for this one, you can see in the summation sequence that the only value changing is the 6/n to 12/n, and eventually to 6n/n.
brazen elbow
Sigma notation means a sum starting from i = 1 to n, so for this one, you can se...
!nosols
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Sigma notation means a sum starting from i = 1 to n, so for this one, you can se...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
limpid mango
thano you so much
brazen elbow
i need answer now cuz its due at 2:10
!noans
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harsh scaffold
help. using the formula \left|x\right|\ ^{\left|y\right|}\left|y\right|^{\left|x\right|}=e^{-\frac{2}{e}} if we input a line at x=-e^{\left(-\frac{2}{e}-1\right)} how would you calculate the intersection point of the 3 lines at y=0.018455508592
desmos link: https://www.desmos.com/calculator/auenv47xgf
harsh scaffold
y=0.018455508592 was found from trial and error
but i need help finding why y intersects at 0.018455508592
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harsh scaffold
also with the inner part of the shape desmos doesnt load all of it but it slowly approaches zero but desmos does not load it, 3rd image is when you click the home button and zoom in using plus. how would i calculate the rate of change, as in if i want to see how tall the curves are at x=0.05 how would that be calculated
also if how would you calculate the complete area under the curves if we have a line at x=-e^((-2/e)-1) all the way to when our 2 lines reach zero
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thick sinew
help. using the formula \left|x\right|\ ^{\left|y\right|}\left|y\right|^{\left|x...
Ys you'll need W-lambert though
|x|^|y| * |y|^|x| = A
I'll just abbreviate terms with letters to not have to write them out as I'm on phone currently
and you're given the line x = B
to find the intersection we can start by inserting into the clover function
|B|^|y| * |y|^|B| = A
e^(ln(|B| * |y|) * |y|^|B| = |A|
e^((ln(|B| * |y|)/|B|) * |y| = (|A|)^(1/|B|)
e^((ln(|B| * |y|)/|B|) * |y| * ln(|B|)/|B| = (|A|)^(1/|B|) * ln(|B|)/|B|
Branching into W0: |y| * ln(|B|)/|B| = W0((|A|)^(1/|B|) * ln(|B|)/|B|)
|y| = W0((|A|)^(1/|B|) * ln(|B|)/|B|) * |B|/ln(|B|)
y = ±W0((|A|)^(1/|B|) * ln(|B|)/|B|) * |B|/ln(|B|)
yields both intersection values for the given line and clover function
the negative term will obviously be the one you saught after in the original question
thick sinew
also with the inner part of the shape desmos doesnt load all of it but it slowly...
"how tall the curves are" is oddly worded, do you refer to the derivative(s) at x=0.05? If so then you'd have to derive the original expression after y and insert for x, which I'll mention before you attempt it, isn't too straightforward
thick sinew
also if how would you calculate the complete area under the curves if we have a ...
This statement isn't too clear either, applying and integrating W-lambert w.r.t. y = -x and y = x gives you the integrals for each symmetrical section of the clover, but you probably refer to something different, mind sketching the integral you seek?
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tender adder
how wrong am i
calm coralBOT
how wrong am i
tender adder
and what other alternative can/should i take
tender adder
these two steps are wrong too, i think
surreal patio
its correct but u wrote it wrong
tender adder
why does x+1 turn into x-1?
ohh yah that minor, i think that will fix when doing in paper (coping)
surreal patio
$$log A - log B = log \big{(}\frac{A}{B}\big{)}$$
tender adder
other everything is right?
potent lotusBOT
JustToPro
tender adder
$$log A - log B = log \big{(}\frac{A}{B}\big{)}$$
no like the way i just removed the logs directly, is that correct?
surreal patio
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tender adder
0 > x^2-4x+2?
tender adder
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tender adder
its not x^2-4x+2 tho
Cuz u multipled x-1 and x-2 when its x+1 and x-2
ohh yah that part 
ill fix it, thankss
.close
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keen grove
Can somebody pls tell me why this is wrong?
calm coralBOT
Can somebody pls tell me why this is wrong?
keen grove
am i meant to take the stationary pts as a point as well?
fair estuary
the mistake in the solution strategy is that it uses the functions values at the stationary points which are not necessarily the left riemann points
fair estuary
yup
also how is 53.579 wrong?
i mean A = A0 + A1 + A2 + A3 = 14.59 + 12.229 + 13.689 + 13.062 = 53.579
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remote mural
help
idk how to start the wuestion and wat the question is asking for
remote mural
and i dont understand the notes that much either 😭 if someone can simplify it-
dull wagon
equation in the notes is wrong
$f(f^{-1}(x)) = f^{-1}(\underbrace{\red{f(x)}}_{\text{not } x}) = x$
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remote mural
**ℝαμΩℕωⅤ**
so just f^-1 (x)?
dull wagon
for what
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remote mural
isnt the answer for this -1? you multiply and divide by b^n and it becomes (a/b)^n and since b > a it tends to zero so it becomes lim x-> infinity 0+1/0-1 which is equal to -1
remote mural
this is what the answer key shows
glacial inlet
b is not greater than a so if you divided it by b^n, it’ll go to infinity and not 0
remote mural
b is not greater than a so if you divided it by b^n, it’ll go to infinity and no...
man im retarted
i read it wrong
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mossy wigeon
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urban relic
and why does it go to -inf?
sharp flower
if you find the limit
at infinity
its 0
not 0
wait
no
it would be 2/49 i believe?
urban relic
what do youknow of the limit of a^x if a is between 0 and 1?
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trail hawk
Can the someone help me with question
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spiral dagger
Can the someone help me with question
let x be the unknown variable
x/4 litres per hour is the draining speed
x/4 - 3 litres per hour is draining speed while it is raining
you can solve it from here?