I think I should have proven that (x,y) is in (A x B) and not in (C x D).
#help-10
dark stirrup
Are you using Microsoft Word btw?
tawdry arrow
Yeah, I have no idea what else to use.
dark stirrup
I recognize the font.
In the equation editor, you can do \eqarray to make an aligned equation array
Add & to where you want to align things, and do Alt+N to go to next line while preserving alignments (pressing Enter will break alignment, a known bug)
Also, a style suggestion, here's how I like to write set equality proofs
tawdry arrow
Thank you man, I just started doing these things digitally. I usually use a pen and paper, but now I have to go digital cause of my uni having everything online.
It's so hard to adapt.
tawdry arrow
Can't keep this open forever, other people need to ask things.
pulsar quarry
this channel is already closed, make a new one
obtuse pebbleBOT
Available help channel!
minor pivot
Hi I was wondering if I could get help with this numerical methods problem
obtuse pebbleBOT
@minor pivot Has your question been resolved?
minor pivot
<@&286206848099549185>
obtuse pebbleBOT
@minor pivot Has your question been resolved?
obtuse pebbleBOT
@minor pivot Has your question been resolved?
minor pivot
<@&286206848099549185>
obtuse pebbleBOT
@minor pivot Has your question been resolved?
junior jay
.reopen
obtuse pebbleBOT
@minor pivot Has your question been resolved?
pale isle
.reopen
royal basin
!status
obtuse pebbleBOT
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
!15m btw
obtuse pebbleBOT
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.
royal basin
you're not supposed to ping helpers unless it has been 15 minutes.
i reminded you about this rule.
i also asked you what progress you made on the problem
royal basin
```
What step are you on?
1. I don't know where to begin.
2. I have begun but go...
right here
ok
so, you know how a clock works, right?
i wasn't implying you were underage.
so then you know that at 6:00 the hour hand points directly at the 6 and the minute hand points directly at the 12
so the angle between them is 180° at that point in time
then as the clock ticks forward the minute hand catches up to the hour hand
so in between 6:00 and catching up, the minute hand will be exactly 90 degrees behind the hour hand.
now let x be the time since 6:00, in minutes.
how many degrees does the minute hand turn through? and how many does the hour hand turn through?
@timid silo
??
what's wrong?
we will be forming an equation in x using the problem and the reasoning i wrote out above, and then solve for x and find the answer.
does this clear up your confusion?
if there's something else you don't understand, please tell me what and i'll explain.
ok... how many degrees does the minute hand turn through in 1 minute?
and in x minutes?
yes
and how many degrees does the hour hand cover in x minutes?
spin, turn, rotate, however else you want to call it.
so, how many degrees does the hour hand turn through in x minutes?
yes correct
so now
so then
the positions of the minute and hour hand, measured as a clockwise angle from the 12-hour mark, at x minutes past the hour are thus:
minute hand = 6x
hour hand = 180 + x/2
do you understand this
at 6:00, the hour hand points at the 6 hour mark.
that's 180 degrees, half a full circle, away from the 12
now write down the equation that says the minute hand lags 90 degrees behind the hour hand.
why 2x...
2/x??
180 + x/2 - 6x = 90, yes this is correct.
now solve for x.
opaque oxide
clever pier
$a=-1-\sqrt{3}i$
warm shaleBOT
ℝage-EMILY
clever pier
I'm assuming this is 'a' based on the image.
opaque oxide
I'm assuming this is 'a' based on the image.
Yeh
clever pier
Yeh
Are you sure of what you took as input for your arctan?
Recheck that.
Now as you know, since sin and cos tend to be both negative in the third quadrant, tan is essentially positive which gives us the value of tan in the first quadrant.
Thus you need to offset to get the correct answer. (sub pi)
Which I'm assuming you've tried in the bottom right.
opaque oxide
hmm
but I thought that if it is in the 3th it should be -pi ?
1th and 4th = + 0
2th = +pi
3th = -pi
clever pier
but I thought that if it is in the 3th it should be -pi ?
1th and 4th = + 0
2th ...
Sorry. My bad. Yeah you're right.
So other than the input you've done everything correctly.
opaque oxide
oh okay thank you !
bdw can you just tell me what input ?
clever pier
Instead of arctan(1/sqrt(3)) it should be arctan(sqrt(3))
a+bi
arctan(b/a)
That's pretty much it. I had a brain fart with the offset thing.
obtuse pebbleBOT
opaque oxide
!close
clever pier
Yep.
opaque oxide
4th times the charm !
obtuse pebbleBOT
Available help channel!
uneven igloo
i did the a part but i couldnt do b c and d
obtuse pebbleBOT
i did the a part but i couldnt do b c and d
uneven igloo
can some1 explain it
obtuse pebbleBOT
@uneven igloo Has your question been resolved?
obtuse pebbleBOT
@uneven igloo Has your question been resolved?
obtuse pebbleBOT
@uneven igloo Has your question been resolved?
lunar nimbus
.
obtuse pebbleBOT
@uneven igloo Has your question been resolved?
undone badger
hello
undone badger
can help with physics numericals?
clear dove
can help with physics numericals?
what level are they? show me the example
pulsar quarry
!show
obtuse pebbleBOT
Show your work, and if possible, explain where you are stuck.
pulsar quarry
i aint going no where
undone badger
i guess book's sol. is wrong
pulsar quarry
mind showing your work?
clear dove
(a)
clear dove
ohh thanks
no problem
undone badger
one more question
clear dove
one more question
yeah?
clear dove
finding information cuz we don't have any of these concepts 😅
undone badger
finding information cuz we don't have any of these concepts 😅
same here
bro leave it
illogical question
another one plz
undone badger
My mind went heated up due to this question
clear dove
My mind went heated up due to this question
same here lol
undone badger
i have another one
clear dove
okay, go ahead
clear dove
yeah, solving rn
undone badger
ok
clear dove
is it possible to have useless info here?
cuz i don't know how to use speed of elevator
wait nvm
undone badger
okk
undone badger
bro i am leaving
undone badger
h,mm
clear dove
T is frictional force
undone badger
okay
bro leave it
actually i was missing one formula
i once again read my book
i got the answer
clear dove
bro leave it
wdym, i already solved that
clear dove
actually i was missing one formula
oh
undone badger
what is numerical value
clear dove
one sec
undone badger
total mass = 680Kg plus 920kg = 1600 kg
clear dove
correct bro
yoo lets go
undone badger
i forgot one formula - Power = Force * Velocity
that is why I was not able to do it
BTW which country?
clear dove
BTW which country?
Belarus
clear dove
I study at university
clear dove
2nd year
clear dove
I am in 10th
btw which country are u from?
clear dove
Are you familiar with IIT - JEE
unfortunately, no
clear dove
this question was from entrance exam of IIT (College)
oh, really?
undone badger
although thanks brother
clear dove
although thanks brother
not at all
clear dove
our central exams have been made more easy recently
undone badger
ohh
clear dove
soo, many students pass exams with high scores
clear dove
bye o/
undone badger
.close
ember island
help
clear dove
- tan(x) (1/cos^2(x) - 1)
2.tan(x) ( (sin^2(x) + cos^2(x) - cos^2(x)) / cos^2(x))
3.tan(x) ( sin^2(x) / cos^2(x) )
ember island
howd u get the first one
clear dove
tan(x) sec^2(x) - tan(x) = tan(x) (sec^2(x) - 1) = tan(x) (1/cos^2(x) - 1)
any questions?
ember island
yeah ion get it hopefully this youtube video helps
rugged barn
1. tan(x) (1/cos^2(x) - 1)
2.tan(x) ( (sin^2(x) + cos^2(x) - cos^2(x)) / cos...
I don’t think you ought to begin with the result that you try to prove
clear dove
I don’t think you ought to begin with the result that you try to prove
wdym, i cannot understand
rugged barn
Even if it’s not really clear, I think they meant to start with tan^3(x) and try to do something with it. Because you can’t start with the result (which is the equality you wrote in 1.) to prove the same result
Nvm
clear dove
i don't think so
obtuse pebbleBOT
@ember island Has your question been resolved?
thick pond
hi i have a quick question
obtuse pebbleBOT
hi i have a quick question
thick pond
kind of a silly one but
in the case of like (XAB)^T where we are taking hte transpose of XAB
is the result X^T * A^T * B^T or B^T * A ^T * X^ T
taking a linear algebra analysis class rn and im trying to do a proof but im unsure about this property
.close
icy ivy
I am learning Sine and Cosine law, how do I know which law to use here since theres two formulas given to me of them
obtuse pebbleBOT
I am learning Sine and Cosine law, how do I know which law to use here since the...
latent walrus
well, what are the requirements for each rule?
icy ivy
For cosine its when we know the length of two sides and the measure of the angle thats like in between themm
and for sine law its when we got two angles and one side
or it can also be two sides and angle that is like the opposite of One of the given sides
so for this we have 1 angle and 2 sides
oh
cosine?
latent walrus
there you go
icy ivy
bro made me use my brain properly
so now I just use this formula right
where it go
just a sec
here
latent walrus
yup
latent walrus
np
obtuse pebbleBOT
@icy ivy Has your question been resolved?
narrow spoke
I'm studying calculus and need help understanding two definitions relating to limits.
obtuse pebbleBOT
I'm studying calculus and need help understanding two definitions relating to li...
narrow spoke
I'll attach pictures from my professor's lecture so that I don't end up botching the notation
The definitions are the following:
I've studied calculus before and I understood it back then, but it's been a while and something about the wording here is really throwing me off. Could someone please help break down what is being established in each definition?
Thank you in advance for the help!
obtuse pebbleBOT
@narrow spoke Has your question been resolved?
worn yoke
if we take a = c - δ and b = c + δ then they are effectively equivalent
the main idea is that if we want the function to equal some number that is very close to L, we can find some range of inputs near x where all of those inputs would create an output in that range around L
say for example, you want to prove that lim(x->4) x² = 16. i could take some ε = 0.01, and challenge you to come up with a δ such that all x that are between 4 - δ < x < 4 + δ, x² must be between 16 - ε < x² < 16 + ε.
you could come back and say, "i pick δ = 0.01. you can check and see that the inequalities hold for that". importantly, you could have picked δ = 0.001, or 0.00006, or all sorts of other δ, and it would still work. but there is at least one δ that works.
also, i can go back and pick another ε. i could say, "ok. now what about ε = 0.00004?" and you could go back and find a δ that works for that. and we'd go on like that forever. in order to prove the limit exists, you'd need to find some expression for δ in terms of ε that would give a valid δ no matter what ε i choose.
narrow spoke
say for example, you want to prove that lim(x->4) x² = 16. i could take some ε =...
so essentially, δ is an expression of a range across the x-axis and ε is an expression of a range across the y-axis, both of which the function (in theory) falls into when we are examining
a limit
sorry, i hit enter early lol
narrow spoke
so essentially, δ is an expression of a range across the x-axis and ε is an expr...
and you're trying to get as close to the value of x = c as possible?
i'm just trying to understand these definitions through wording that falls more neatly into place
so please let me know if i'm missing anything
worn yoke
here's a diagram of it (in this case ε = 0.1, but it can be anything)
narrow spoke
here's a diagram of it (in this case ε = 0.1, but it can be anything)
thank you!
i think this makes more sense now
.close
rotund halo
obtuse pebbleBOT
rotund halo
solve the inquality
narrow spoke
i think you can start off by "multiplying" everything by the log function
fathom flicker
The kid from Akron 👑
narrow spoke
so you will get 2xlog(5) - (x+1)log(6.5) > log(-125)
narrow spoke
i would rewrite this: 2xlog(5) - (x+1)log(6.5) > log(-125) as
2xlog(5) - (xlog(6.5)+log(6.5)) > log(-125)
so then you get 2xlog(5) - xlog(6.5) - log(6.5) > log(-125)
you're solving for x, so you isolate the log expressions with x
2xlog(5) - xlog(6.5) > log(-125) + log(6.5)
what do you think you're going to do next?
rotund halo
idk im new to it
narrow spoke
now that you have all the x terms on one side
narrow spoke
idk im new to it
no worries
since you're solving for x, you want to keep isolating until you get an inequality that expresses what x is
so what i would do is factor x out
narrow spoke
2xlog(5) - xlog(6.5) > log(-125) + log(6.5)
so you get x(2log(5) - log(6.5)) > log(-125) + log(6.5)
what do you think might be the next step?
it's okay if you don't know
rotund halo
idk\
narrow spoke
btw, it does say -125 in the original inequality, right?
rotund halo
yes
narrow spoke
so you get x(2log(5) - log(6.5)) > log(-125) + log(6.5)
the next step for this is to continue isolating x
i like to use the expression 2 x 3 = 6 to illustrate how
narrow spoke
i like to use the expression 2 x 3 = 6 to illustrate how
@rotund halo how do you get 2 from this expression?
the reason i ask is because if you think about it, the inequality that you're currently looking at in this step has a similar structure
(2) × (3) = 6
and
(x) × (2log(5) - log(6.5)) > (log(-125) + log(6.5))
pliant plume
W name lebron
rotund halo
<@386962123710005251> how do you get 2 from this expression?
divide by 3
narrow spoke
divide by 3
yep, divide 6 by 3
narrow spoke
(2) × (3) = 6
and
(x) × (2log(5) - log(6.5)) > (log(-125) + log(6.5))
so keeping this framework in mind, how would you get x from our current expression?
narrow spoke
i didnt learn this
gotcha
i actually want to backtrack really quickly because i realized i made a mistake
the reason i asked about -125 is because a log function's domain is 0 and all positive numbers onwards
so i'm going to rewrite step 1 a little
rotund halo
alright
narrow spoke
i feel like there's a better way to solve this, so maybe once i finish explaining ping the helpers role to make sure everything is in order
but what i would do is this: divide everything by -1
narrow spoke
that way we get 6.5^(x+1) - 5^(2x) < 125
then we follow the same steps i outlined earlier
and we end up with the following
(x+1)log(6.5) - (2x)log(5) < log(125)
(x)log(6.5) + log(6.5) - (2x)log(5) < log(125)
(x)log(6.5) - (2x)log(5) < log(125) - log(6.5)
(x) × (log(6.5) - (2)log(5)) < (log(125) - log(6.5))
isolate x, as originally planned
using the 2 x 3 = 6 -> 6/3 = 2 framework, we get
x < (log(125) - log(6.5))/(log(6.5) - (2)log(5))
does this make sense?
narrow spoke
yes, you can solve the log expressions for numbers
since all the "x" values within them are positive
rotund halo
can you show me the process of simplifying it however much it can be and tell me what answer you get
narrow spoke
sure
narrow spoke
can you show me the process of simplifying it however much it can be and tell me...
i'm going to be honest, i'm not quite sure how to simplify it. atp i would simply plug everything into a calculator. maybe you could ping the @/helpers role and ask if we're approaching this inequality correctly?
maybe there is a way to simplify it that i'm not seeing
narrow spoke
sorry, i hope i was useful nonetheless!
rotund halo
do you use log(a) - log(b) = log(a/b) and 2*log(c) = log(c^2) formulas
narrow spoke
do you use log(a) - log(b) = log(a/b) and 2*log(c) = log(c^2) formulas
actually yeah you do 😭
rotund halo
is that what you were missing
narrow spoke
yeah lowkey forgot abt that
narrow spoke
i'm not sure how to simplify this further, though the formulas might help
sorry, i think i'm out of my depth. i need to review all this
<@&286206848099549185>
obtuse pebbleBOT
@rotund halo Has your question been resolved?
obtuse pebbleBOT
@rotund halo Has your question been resolved?
last gorge
Question: Grading only the difficult questions, the researchers found that scores on the blue exam had a distribution with a mean of 53% and a standard deviation of 15%, while scores on the red exam had a distribution with a mean of 39% and a standard deviation of 12%. Assuming that both distributions are approximately normal, on which exam is a student more likely to score below 20% on the difficult questions, the blue one or the red one?
obtuse pebbleBOT
Question: Grading only the difficult questions, the researchers found that scor...
last gorge
heres my work, i think the answer is red because probability is bigger
but i check my work on chatgpt, it says it is blue because of the bigger abs value of z score
the 2.2 should be -2.2
I don't know if im right or gpt is right
<@&286206848099549185>
obtuse pebbleBOT
@last gorge Has your question been resolved?
neon remnant
If I have 2(×-1)(×-3) then the zeroes of the function would be 1 and 3 right?
neon remnant
There would be no zero at 2 right
timid silo
yes
azure anchor
Is this enough for an informal argument?
obtuse pebbleBOT
Is this enough for an informal argument?
azure anchor
My logic is thinking about Nul A containing more than {0} when det A = 0, so there must be a "loss" of dimension which would make it impossible for A to span R^3, but some plane or line would generate instead.
Rank A < n, where n = dim A
.close
misty ridge
Please idk how this work
obtuse pebbleBOT
Please idk how this work
daring sorrel
Please idk how this work
What is degrees of angle EBD
open whale
ok so basically you know that
ABC = 180 degrees
And you know that BE bisects ABD
So basically what this means is that
ABD = 2 * (ABE) = 2*(2x+10)
so you know that 2(2x+10) +3x+20 = 180
Now solve for x
open whale
Because the word bisects means it is exactly in the middle
misty ridge
why do we time 2 in the middle
open whale
It splits the angle into 2 equal angles
misty ridge
OHHHHHH
open whale
Ill give you an example
If your angle is 90 degrees, and then you bisect it
you get 2 angles
45 degrees each
Bisect means you split the angle in half
so ABE = EBD
because BE bisects ABD
open whale
no worries
misty ridge
.close
timid silo
obtuse pebbleBOT
timid silo
why is this wrong
meager aspen
, w solve 5+47x+10(1-x^2)=0
warm shaleBOT
meager aspen
Why do you think it's wrong? Answer isn't matching?
timid silo
Why do you think it's wrong? Answer isn't matching?
they told me should be basic angle and no negatives
i got -17.457
-0.304.. if rad
meager aspen
they told me should be basic angle and no negatives
The bounds on the angle might be [0, 2pi]
meager aspen
Sin is negative in [pi, 2pi]
timid silo
Sin is negative in [pi, 2pi]
how to get the corretct ans
meager aspen
, w arcsin(-3/10)
warm shaleBOT
timid silo
, w arcsin(-3/10)
still neg
meager aspen
, w sin(pi+0.30469)
warm shaleBOT
meager aspen
At last
dry grail
sin(pie + theta) = -sintheta
sin(2pie - theta) = -sintheta
@timid silo
meager aspen
Add pi to your result
meager aspen
, w sin(2pi-0.30469)
warm shaleBOT
timid silo
https://tenor.com/view/yoda-there-is-another-star-wars-jedi-gif-5140031
alright tyyyyy
i goot 162... and 342...
timid silo
sin(pie + theta) = -sintheta
sin(2pie - theta) = -sintheta
<@456226577798135808...
tyyy
.close
flat niche
I want to know if an integral of a rational function is always a rational function. I used integral of 1/x as an example for why this is false but idk if thats adequate. is it?
obtuse pebbleBOT
I want to know if an integral of a rational function is always a rational functi...
meager aspen
The proof becomes invalid whenever you find one example the proof doesn't cover/prove
So yeah that counterexample is completely correct and adequate
flat niche
sick
flat niche
makes sense tho lol. i mean its in base e so of course its gonna spit out a silly number
meager aspen
Yup
flat niche
ima pull up another 2 rq
i want to prove that cosx to an odd power is always equal to zero
meager aspen
cos^n(x) is always zero where n is odd
flat niche
with the example of the integral of (cosx)^2m+1
brave bramble
Is that over any certain bound?
meager aspen
with the example of the integral of (cosx)^2m+1
Any bounds on the integral
flat niche
how am i supposed to justify that tho
meager aspen
A) cos^(2m+1) (x) is not zero for all x
The integral is different case
I will give you an example
flat niche
yeah since the areas
meager aspen
Exactly
flat niche
ur always getting an equal positive area as the negative area so it always cancels out
flat niche
but im just wondering why its like that for odds
meager aspen
Oh
Because (-ve number)^even is positive
While (-ve number)^odd is negative
You need some negative points for the graph to go below x axis or say have negative areas
flat niche
whats the -ve number mean
meager aspen
The range of cos is from (-1,1) right?
[-1,1]
The range of even power of cos will always be [0, 1]
flat niche
whys that?
meager aspen
Nah
flat niche
lol overcomplicated that one haha
ok last one is just asking if the integral of the sqr(1+x^2) from 1 to 2 has a finite real value which it def doesnt right
because its a semi circle and the bounds end at 1 and -1
obtuse pebbleBOT
Available help channel!
midnight steppe
How could I show that S is closed under * here?
obtuse pebbleBOT
How could I show that S is closed under * here?
sage geode
First state why xy is also a real number and use closure of R under +
spice citrus
Also what happens if x * y were equal to -1?
sage geode
Ah yeah and that
midnight steppe
If x * y = -1 then exactly one of x and y is negative?
Yeah I’m mainly stuck up on the x!=-1
junior kelp
what could happen when "x star y == -1"?
kind hawk
if n is fixed then G is a finite group
kind hawk
so (a) cant be true
royal basin
cause im not 100% on that
kind hawk
try finding all the elements for GL(2,Z8) explicitly
royal basin
wait, what even is a self-conjugate element
is it an element that's literally conjugate to itself?
cause that would apply to every element of G
what am i missing
kind hawk
I would have guessed fixed under every conjugation
but yes good question didnt really think about it
royal basin
ah, ONLY of itself.
so then for all h ∈ G you must have hgh^-1 = g
ok yeah then thats equiv to being in the center
and n > 1, yes?
right
hm
kind hawk
sry I meant only the self conjugate ones
kind hawk
so the zero matrix is in the set?
is the zero matrix in G?
what about the matrix with 2s on the diagonal
in a ring the claim that matrix is invertible iff det is nonzero is false
the correct claim is: matrix is invertible iff the determinant is invertible
or instead of arguing by determinant you can also argue directly by showing linear (in)dependence or not
in what world is Z_8 not a ring
royal basin
how have you not started with rings yet the ring Z8 is staring you in the face
kind hawk
why not just limit yourself completely and forget what matrices are cause thats linear algebra and not group theory
I am not asking you to learn all of ring theory
but some fundamental results about matrices over a ring could be good
the point is, $\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} 4 \ 4\end{pmatrix} = 0$
warm shaleBOT
Denascite
kind hawk
so the matrix is not invertible
royal basin
dena's point is 2I does not belong to your group at all
because it fails to be invertible
obtuse pebbleBOT
obtuse pebbleBOT
✅
kind hawk
it includes 1/det(A) as a factor
aka the inverse of det(A)
which therefore has to exist
royal basin
what's the inverse of 4 in Z_8?
kind hawk
"nonzero" and "has inverse" are not the same in a ring anymore
thorny kite
I want to know if this is wrong
obtuse pebbleBOT
I want to know if this is wrong
thorny kite
the question is what happens if λ ∈ [0, 1]
the u and v values are right
also this is a vector problem
alpine raven
what is the task ?
thorny kite
graph λu + v
thorny kite
the question is what happens if `λ ∈ [0, 1]`
.
the resolution is not made by me but I don't have the confirmed correct answer
but my intuition tells me it is wrong
but I dont know
because I don't know what λ ∈ [0, 1] implies in the vectors
I tried searching
couldn't find and ended up here
alpine raven
if λ = 0 , λu+v = v
if λ = 1 , λu+v = u+v
now for all values of lambda in (0,1), I dont really know how to tell what happens
thorny kite
what would "all values of lambda in (0,1)" mean?
thorny kite
in this case (0,1) is a vector or a point?
thorny kite
im not american I don't know what a set would be
alpine raven
but you are working with [0,1] which is a set...
thorny kite
ok i know now xd
alpine raven
(0,1) is all the values between 0 and 1 (0 and 1 not included)
[0,1] is all the values between 0 and 1 (0 and 1 included)
alpine raven
yes
thorny kite
so it would be what happens while lambda is between 0 and 1
0 and 1 are included tho
in the task
alpine raven
yes
thorny kite
I want to know if this is wrong
what I don't understand is why this dude added up the points
I thought the first one was the vector
and the second one was a point in a line with the angle of that vector
am I right??
quaint plume
Ye I'm pretty sure
alpine raven
u and v are both vectors
thorny kite
then 0,0 should be in the line right?
if a vector is just one point then it starts at 0,0 right?
and ends at the point
alpine raven
if a vector is just one point then it starts at 0,0 right?
a vector cant be a point only (unless its the zero vector idk)
thorny kite
im pretty sure that if a vector gets defined by a point it starts at 0,0
is that wrong?
alpine raven
im pretty sure that if a vector gets defined by a point it starts at 0,0
how can u define a vector by a point ?
thorny kite
is it wrong?
alpine raven
is it wrong?
I dont know, im asking lol
thorny kite
i mean if you can assume 0,0 is a point of the vector then yes
you get 2 points
so you can define a vector
<@&286206848099549185>
obtuse pebbleBOT
@thorny kite Has your question been resolved?
obtuse pebbleBOT
@thorny kite Has your question been resolved?
obtuse pebbleBOT
@thorny kite Has your question been resolved?
thorny kite
obtuse pebbleBOT
@thorny kite Has your question been resolved?
obtuse pebbleBOT
@thorny kite Has your question been resolved?
timid silo
what do you have to do
obtuse pebbleBOT
@thorny kite Has your question been resolved?
obtuse pebbleBOT
@thorny kite Has your question been resolved?
steel goblet
obtuse pebbleBOT
steel goblet
i how no idea how to graph a multivariable function
dark stirrup
You're only asked to graph the domain
That is, all valid (x, y) that f(x, y) can operate on.
obtuse pebbleBOT
@steel goblet Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
west heron
obtuse pebbleBOT
west heron
For electric potential at let’s say point P do I use kqr for each charge then add them? Do the angles play a role?
static beacon
For electric potential at let’s say point P do I use kqr for each charge then ad...
angles do not matter as long as you have the exact distance from charges to tthat point
round tundra
obtuse pebbleBOT
round tundra
I need to find the oblique asymptotes to this function
I've found the slope of them but can't seem to get the constant
as in I'm trying to find a and b but I've found a through fucky means
b always seems to be the wrong degree however I rearrange the equation
<@&286206848099549185>
flint hound
Yes?
round tundra
?
round tundra
can I ping again?
flint hound
Sure.
clever ember
shoot hmm
round tundra
ah
this one is annoying enough where wolfram alpha won't even give me the correct answer
clever ember
who's that lol
round tundra
online calculator
very useful
usually
that is what I've found so far
this should be pretty solvable
it clearly has a graphical solution
just having a a hard time finding the exact value of b
clever ember
are you taking the limit of the whole thing or just the first part of the function
idk if that changes anyting
clever ember
wait no that wouldn't change anything
because it's an asymptote
right?
so as x approaches infinity we'd get the asymptote(?)
round tundra
yes
that's the idea
except the asymptote is a function
or line
which is ax+b where a is pi/8
just can't find b
round tundra
that is how I would do it if there wasn't a trig function in there
clever ember
yea that's where i'm getting tripped up
hmmm
shoot
maybe this will work?
audio is awful tho
clever ember
did it work?
round tundra
no idea
clever ember
i didn't watch the full video
round tundra
but this is what I'm trying to do
round tundra
I mean he perfectly described the method I tried to use
fathom flicker
please don't tag me randomly
round tundra
sorry
obtuse pebbleBOT
@round tundra Has your question been resolved?
ember venture
help please
obtuse pebbleBOT
help please
timid silo
well id first suggest writing in the given measures
ember venture
answer is in degrees, is 36 right?
timid silo
ill let someone else figure that out, sorry didnt realize you just needed confirmation
too tired to do it
royal burrow
I have a rotation about the x and y axes
obtuse pebbleBOT
I have a rotation about the x and y axes
royal burrow
how do I convert this rotation to a vector in 3d?
so for e.g rotX = 30deg, rotY = 5deg
how do I get an x,y,z
would combining the bottom two operations possibly work?
(nvm the library im working with already had a builtin method)
gleaming rose
A mole of ideal monatomic gas at 0°C
and 1.00 atm is warmed up to expand isobarically to triple its volume. How much heat is transferred during the process?
obtuse pebbleBOT
A mole of ideal monatomic gas at 0°C
and 1.00 atm is warmed up to expand isob...
@gleaming rose Has your question been resolved?
fathom flicker
obtuse pebbleBOT
fathom flicker
Was wondering if anyone could give me feedback on my strategy for this.
I'm thinking for 1000! to end with 249 zeroes it must contain a factor of 10^(249) and no larger/smaller on the power.
so, 1000!'s prime factorization must contain exactly 5^(249) and 2^(249)
so I'm thinking I might want to show 5^(249) divides 1000! and 5^(250) does not divide 1000! and likewise for 2^(249)/2^(250)
royal basin
almost but not quite
fathom flicker
would that work?
oh because it contain another factor of 5 without a 2
and still end with same zeroes
royal basin
or a two without a five.
royal basin
of the exponents on 2 and on 5 in the prime factorization of 1000!, you would merely need the LOWER of the two to be 249, not both of them.
fathom flicker
great yeah that is a fix
and then I am new to number theory
I am thinking this might use Fermat's little theorem?
it seems like the only useful one I have
royal basin
for what purpose
fathom flicker
I don't know honestly, I have like 5 results from lecture and it seems the most relevant
royal basin
you don't need fermat's little thm here you just need to work out the prime factorization of 1000!
fathom flicker
oh really?
I thought that would be super hard
I can just try though
I haven't yet
royal basin
no, it's not hard.
fathom flicker
how do you do that without finding all the other prime factors though?
yeah actually it does still seem difficult but maybe I am going about it the wrong way
here is what I would think to do
1000*999*998*.....*3*2*1
1000=2^(3)5^(3)
999=3*333=3^(2)*111=3^(3)*37
and continue like that
but if you are saying it is not hard
then this must be the wrong approach
I assume
fathom flicker
all the primes in between 1 and 1000?
royal basin
yes
but you're interested in 2 and 5 only
think: which numbers give you a factor of 2?
fathom flicker
all even numbers
royal basin
which numbers give two factors of 2? which ones give three and so on
fathom flicker
hm okay I think I see what you mean
and then since we are using all of the factors
we end up picking up all these factors
oaky
I'll work on that
ty Ann
also, do you know what Fermat's little theorem is used for?
I have that
a^(p-1) congruent 1 mod p
and I don't really know modular arithmetic super well
royal basin
computationally? finding remainders of high powers
fathom flicker
just wondering like why I might have been given it
for what types of problems. because I see prime and I think of that theorem
since it involves primes
remainders of high powers
so something maybe like #3
might use it
these are my assigned problems
royal basin
idt you will need fermat's little thm for any of these
fathom flicker
ah okay then
I'll forget about it for now
gonna go see what I can do with the prime factorization
ty again
obtuse pebbleBOT
@fathom flicker Has your question been resolved?
fathom flicker
which numbers give _two_ factors of 2? which ones give _three_ and so on
okay so this isn't going as smoothly as I thought it would
so I wrote out like what is possible
2,4,8,16,32,64,128,256,512
and I wanted to know like which numbers had those as factors
because those will give me the various powers of 2's?
royal basin
not the best
fathom flicker
so like
512
256/512/768
128/256/384/512/640/768/896
and as you can see
not good
royal basin
there are 500 even numbers in the product that give us at least one factor of two
fathom flicker
yes
royal basin
but some of these give us more than one factor of two
fathom flicker
yes
royal basin
how many give at least two each?
fathom flicker
all numbers divisible by 4
royal basin
right, and how many of those are there
fathom flicker
250?
fathom flicker
Okay I tried this at first
but I got confused at the step after this
125
by the same logic
and then we get a fraction
royal basin
let's take this to its conclusion
no, we don't.
how many multiples of 16 are there under 1000?
warm shaleBOT
fathom flicker
so is it just 62 then
fathom flicker
why do we know that rounding down works
what does it mean to have half an integer that is divisible by 16
royal basin
it does not
fathom flicker
this is why I was confused about how we are certain there are 250 numbers less than 1000 that are divisble by 4
royal basin
the number of multiples of m not exceeding n is floor(n/m) always
royal basin
this is why I was confused about how we are certain there are 250 numbers less t...
1*4, 2*4, ..., 250*4
likewise 1*8, 2*8, ..., 125*8
fathom flicker
okay I do see now
so there are 500 div by 2
250 div by 4
125 div by 8
62 div by 16
31 div by 32
15 div by 64
7 div by 128
3 div by 256
1 div by 512
royal basin
yes
fathom flicker
that is what I was getting from my table
but listing them out
was not the good method
royal basin
and in fact now to find the total exponent on 2 you can just add all these counts together
fathom flicker
but isn't 512 both one of the numbers that is divisible by 512 and divisible by all the rest
so are we not double counting
or more than double
royal basin
yes it is, and yes we are, just as we're supposed to!
512 gives us 2^9, and we're counting it 9 times.
fathom flicker
oh since we actually are multiplying by all of these
we get them every time
494
powers of 2
fathom flicker
indeed
warm shaleBOT
Result:
994
994
fathom flicker
that's a lot of 2's
royal basin
now calculate v_5(1000!) in the same fashion
fathom flicker
2^(994) is in 1000!
prime factorization
we can say
okay and now I see how to do it for 5 aswell
what I wrote is
1000! ends with 249 zeroes if and only if its prime factorization contains 2^(a)5^(b) where either
a=249 b>=249
b=249 a>=249
is that true
I think so
and now we are going to show that b=249 and a=994>249 so it works out
unless it is actually a lie that 1000! ends with that many zeroes
in which case I will deduce that
royal basin
where either
a=249 b>=249
b=249 a>=249
or to put that more concisely, min(a,b) = 249
fathom flicker
oh that works much better
royal basin
it will be.
fathom flicker
indeed
200 with 5^1
40 with 5^2
8 with 5^3
1 with 5^4
249 total
done
viola
ty Ann
.close
storm belfry
Let θ,φ∈[0,2π] be such that
2cosθ(1−sinφ)=sin^2θ(tan θ/2 +cot θ/2)cosφ−1
tan(2π−θ)>0 and −1<sinθ< − root3/2
obtuse pebbleBOT
Let θ,φ∈[0,2π] be such that
2cosθ(1−sinφ)=sin^2θ(tan θ/2 +cot θ/2)cosφ−1
tan(2π...
storm belfry
i want to find the range of φ
i have simplified it to 2cosθ + 1 = 2 sin (θ+φ)
with the given conditions, i found that θ belongs to (3pi/2, 5pi/6)
and i found that 1/2 < sin (θ+φ) < 1
obtuse pebbleBOT
@storm belfry Has your question been resolved?
storm belfry
<@&286206848099549185>
obtuse pebbleBOT
@storm belfry Has your question been resolved?
brave plume
log(4^x+5^x)
timid silo
there is nothing to solve