yes, this is also a valid proof !
#help-38
wraith hinge
yes?
nova spire
No, the idea is to show that such an epsilon exists for ANY x
"for all x"
"exists epsilon"
If I pick a random x, I only need you to provide me with a SINGLE epsilon that works
nova spire
you could have also gotten intuition from a graph, lemme send it
When you're tasked to prove a statement that starts with "There exists epsilon...", you don't need to find ALL the epsilon that will work, we've only asked you to find at least one so why make life more difficult?
wraith hinge
i suppose
and so you've drawn a box
to represent it goes from (-1,1) on both
and its stripped lines because an open interval
nova spire
yes, the circle should be stripped too
wraith hinge
what is the circle for
nova spire
what is the circle for
B((x,y),epsilon)
nova spire
and why is it a circle
In 2D, open balls are... disks (with respect to the regular norm)
and in 3D, open balls are... balls
if you recall, the usual norm in $R^2$ is $|(x,y)| = \sqrt{x^2+y^2}$.
And so if you draw $B((x,y),\varepsilon)$, it's the points $(x',y')$ such that $(x-x')^2+(y-y')^2 < \varepsilon^2$, so just the inside of a circle
solid kilnBOT
rafilou2003
nova spire
No, what unfortunately everyone fails to mention is that in the end, we have to prove our conditions for (x',y') in our ball : |x'| < 1 and |y'| < 1
this is done in the example I gave you
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wraith hinge
here
Could you write it a LaTeX or on paper?
@nova spire
and where does it originate from
We introduce epsilon = min(1-|x|,1-|y|) > 0. is what my question is about
nova spire
We introduce epsilon = min(1-|x|,1-|y|) > 0. is what my question is about
We work our way backwards, we would like in the end to have |x'| < 1 and |y'| < 1.
But we know that |x'| < |x| + epsilon and |y'| < |y| + epsilon (triangular inequality).
In order for this reasoning to work, we need |x| + epsilon <= 1 and |y| + epsilon <= 1.
Thus we need to have epsilon <= ... and so we take an epsilon that verifies this
So to recap :
In order to prove the "epsilon neighborhood" definition, we need to find, for each point x in the supposed open set, AN epsilon that gives us a valid epsilon neighborhood.
To find this epsilon, we work our way backwards, knowing that we want to prove in the end that .... (depends on the conditions of your set).
This allows us to find at least one epsilon, that we use to prove the validity of our epsilon neighborhood
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midnight bloom
if f = x + log x and g = log x is f equal or not equal to O(G)?
midnight bloom
ah sorry wait wrong given
is this f equal or not equal to O(g)?
big O
notation
cause if im solving this
i would look at the left hand side which is x + log and just get the bigger term which is x
then compare that with the log x
is that how i should do it
or should i consider both the x + log in the lefthandside vs the log x at the right?
n ot sure which is the right way
not entirely but is f is equal or not equal to O(g)?
kinda confused cause idk if my teacher is right so
so would it be a yes or a no? for the O(g) only
icl?
<@&286206848099549185>
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wraith hinge
Hi, how do I go about this?
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
y = 1 - x^2, y = 0; about the x-axis
turbid blaze
help what is 2.34-5.3439
stoic iron
help what is 2.34-5.3439
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turbid blaze
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zinc ginkgo
Hi, how do I go about this?
Find the volume of the solid obtained by rotating t...
Did you draw the picture
wraith hinge
Yeah, I did
wraith hinge
Did you draw the picture
Well, I drew the 2 lines
The rotation is confusing
To represent
zinc ginkgo
The rotation is confusing
wraith hinge
https://tutorial.math.lamar.edu/classes/calci/volumewithrings.aspx
How do I represent this in 3D
wraith hinge
There's a picture in the link
Is the volume I'm looking at a perfect sphere?
wraith hinge
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sharp cosmos
Here's what I've tried so far:
any help?
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any help?
sharp cosmos
I created an augmented matrix, but I don't know what that actually says about the span of the three vectors at all
unique minnow
Isn't the zero vector in the span of any vector by letting the combination coefficients all be 0
?
unique minnow
Or if you think of the span as the smallest subspace containing those three vectors, then the subspace must contain the zero vector
real saddle
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real saddle
why is this wrong?
i do
2a = 8
a = 4
4^2 - b^2 = 3^2
-b^2 = -7
b^2 = 7
to find b
and since the minor is vertical its horizontal so its x/a + y/b = 1 i think
oh wait thats the length of the minor isnt it
but wouldnt that still be the same
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gaunt pagoda
how is (1/4)/(sqrt(3)/4) = sqrt3/4????
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how is (1/4)/(sqrt(3)/4) = sqrt3/4????
wind storm
in the right triangle RST, median SU is drawn to hypotenuse RT. If SU = y + 5, RU = 3y+2+x and UT = x+2y-1, find the value of length SU
wind storm
how would i go about this
the answer in the answer key is 2 but idk how my teacher got there
nvm lol
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somber spire
is this correct
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is this correct
sturdy heron
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winged kestrel
high heron
part (a) is indeed just asking for the area of a sector with radius 2 and angle pi/6, yes
and that is pi/3
(unless I've also messed up)
winged kestrel
alright, and one question, how would I go about finding how fast the area changes at t=1? Would I fill in t=1 into the derivative of the area formula with the formulas for r and theta in it?
so differentiate with respect to t is what I mean basically
high heron
yep
analog tapir
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analog tapir
how would i solve this
i gotta solve for acceleration in the system n then solve for the tension
force of gravity = mass of C (5.0kg) times acceleration (10 m/s^2 or gravity)
force of gravity = 50N
and then use fg =ma to solve for acceleration
50n = mass of the system (11kg) times a
50/11=4.5 acceleration is 4.5 m/s^2
then do ft-fg=ma
ft-50=11*4.5
ft=99.5
but my answer key says the right answer is 18
how is 18 the right answer
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analog tapir
oh bet ty
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analog tapir
nah :3
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knotty locust
i hate physics
@kindred pier
kindred pier
i hate physics
Physics is a good life
thin thorn
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thin thorn
I need help for question (b). I don’t understand the question so I drew a few sketches. The blue “sin theta….” is from the marking scheme, which I don’t quite get how they got this
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sleek aurora
I take it that P_3 are polynomials of degree 3 with real coefficients?
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blazing panther
hey can anyone write for me inequality cauchy
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hey can anyone write for me inequality cauchy
blazing panther
thanks
glossy root
Are you referring to the Cauchy-Schwarz inequality?
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glossy root
👍
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bright lion
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bright lion
Oop sorry there's an error its this
This is my answer
If someone's free please help lol it looks more complicated than it is, I'd like to check my work
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zinc ginkgo
,rotate
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bright lion
Wait sorry I ended up with this after some fixing
Here's the existing albeit terrible working
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placid grail
The actual question is to show that rk A + rk B <= rk AB which turns into dim ker A + dim ker B >= dim ker AB
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glad tundra
What is the domain and end behavior for this function?
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What is the domain and end behavior for this function?
glad tundra
Yea would I write it like (-∞,-2)U(1, ∞)?
full hull
what about -1
nimble stone
is that x(x-2)(x+2)?
glad tundra
Sorry x(x-1)(x+2)
nimble stone
ah okay, your domain seems sensible so far
however
as seif said, what about the bits between -2 and 1
glad tundra
Would it be like (-2,0)U(0,1)
full hull
find all the values where the under part is equal to 0 then exclude them from R
nimble stone
$\mathbb{R}\setminus {-2,0,1}$
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AℤØ
glad tundra
How would I go about the end behavior ?
nimble stone
find the limits as x tends to + or - infinity
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lofty arch
my brains having a fart
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lofty arch
infinite sum (n-1)/n^2 diverges
how i use comparison test
ik how to use the test this question is just fucking my brain
unreal loom
why not
lofty arch
if its larger than 1/n then n^2-1/n^2 is greater than 1
unreal loom
but the limit of the ratio is 1 is it not?
high heron
n-1 < n so n-1/n^2 < n/n^2 = 1/n
lofty arch
mabye like 1/sqrt(n)
lofty arch
but the limit of the ratio is 1 is it not?
that says nothing about convergence i thought
unreal loom
wait are you using limit comparison test or comparison test
lofty arch
im showing it diverges so comnparison test
high heron
it's less than 1/n only by just a bit
(I'm going to talk about intuition a bit here) so if you shrink 1/n correctly, you're going to be able to still get a divergent series below yours
lofty arch
yes as 1/n is divergent if you make it smaller by a bit it will still b divergent
high heron
sqrt(n) < n so actually 1/sqrt n is bigger
unreal loom
hint: ||you dont need to change the power of n||
unreal loom
yes
lofty arch
hell yeah
ty
jeez
dk why i didnt think of that
been studying all day for test tomorow o
high heron
in general, if you have a series that's basically something else
like here it's basically n/n^2 because you can ignore the 1 when n gets really large
lofty arch
so shkld compare with that series
high heron
scaling by a factor usually works to get something on the side you want (greater for convergence, less for divergence) for n large enough
lofty arch
nd pop a constant on that bad boy if it aint working yet
high heron
(similar example, (x+1)/x^3 is greater than 1/x^2 but not than 2/x^2)
lofty arch
i very much see this was annoying me cus of how easy it was nd i was easily doing much harder ones ahahaha
lofty arch
(similar example, (x+1)/x^3 is greater than 1/x^2 but not than 2/x^2)
yesss just worked that out ty so much ill keep this in mind very helpful :))))
high heron
I'm not sure if there's a general rule you can use, or if it's even true in general, but I think it's helpful for a quick guess
lofty arch
yeh thats what analysis be all about guessing and loosey goosey approxikmations to get ur idea to work
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high heron
can you write the question down? or link the timestamp of the exact question at least
ok so the area gained is $(dx)\cdot\left(\frac1x\right)$, and the area that was lost is $x\cdot d\left(\frac1x\right)$. How would you express that the total area doesn't change (because it stays fixed at 1)
solid kilnBOT
Edward II
high heron
the important thing isn't that the area is 1, it's that it doesn't change, and those expressions (given in the video) represent changes in area
what's 'it'
I mean, yes, but what does that have to do with the area
yes
so how would you express that the area is not changing, using the fact that we have expressions (given in the video) for the area of the extra area on the right of the rectangle, and the lost area on top of the rectangle
ok nvm I don't think the expressions are given, for some reason I thought they were
but they're areas of rectangles with side lengths that are given in the video
yes
yes
I was thinking of a different way but this works too
yes
oh wait I don't think it will
ok
what's the area of the rectangle added on the right when increasing x by dx
ignore the fact that the area is constant for now
just the area of that rectangle
what's the area of the rectangle removed on the top
how are these two related?
we've added one, removed the other, and ended up with no change
yep
except we throw the - into the d(1/x)
no as in
consider d(1/x) as negative so that this is inherently negative and you don't need to put an extra minus to say that we're removing it
I'm not sure how to explain this bit intuitively
grant comments on this as well
it means instead of thinking of the top rectangle as being removed, think of it as adding a negative rectangle
so instead of this being subtracted, it's being added instead
and we're encoding that inside the d(1/x)
alternatively, that the rectangle being removed has area -d(1/x) * x because d(1/x) is negative so we need that - to make length make sense
more like -d(1/x) is the actual length of the side
rather than just d(1/x)
no I'm saying to think of d(1/x) as negative, so for it to represent a length (which is positive) we need to say the side length is -d(1/x), so the area subtracted is -x*d(1/x)
so the area lost is now -x*d(1/x), gained is dx*1/x, and so the overall formula
is (dx * 1/x) + (d(1/x) * x) = 0
where the + is from two negatives cancelling out
yep
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digital ridge
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digital ridge
am i doing it right?
vagrant marsh
You forgot to add the area between 0 and 4/3
vagrant marsh
Yeah
Youre kissing a parenthesis in the last line when you plug in -2 but otherwise looks fine
digital ridge
wait wdym>
digital ridge
ahh aight
thank you
wait for the last line
its not meant to be x
but k
right?
at the end
4/3k and 0k
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grave barn
is this channel available
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is this channel available
sleek talon
grave barn
wraith hinge
,rccw
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grave barn
The radius of Circle A is three feet less than twice the diameter of Circle B. If the sum of the diameters of both circles is 49 feet, find the area and circumference of Circle A.
<@&286206848099549185>
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please help me i dont understand
PLEASE SOMEONEEEE
okay you know what never mind
ugh
i'll figure it out elsewhere
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fierce cloak
could you explain how this equation was simplified?
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could you explain how this equation was simplified?
fierce cloak
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shrewd ridge
nimble stone
you can open this one again if you want
wraith hinge
oh oops i thought the helpers were bots...
full dock
oh oops i thought the helpers were bots...
F
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hallow crane
Anyone know how would i do this?
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Anyone know how would i do this?
hallow crane
If 8 players each play every other player then i'm guessing there's 7! ways already
But my options to select from doesn't really deal with such big numbers:
My options are:
16
32
35
128
256
8-1
8-2
8-3
8-4
8-5
8-6
8-7
then 7-6,7-5,7-4,7-3,7-2,7-1
oh wait it should be
7+6+5+4+3+2+1
sleek aurora
That's correct
sleek aurora
Now half the players are eliminated, so in the next stage only 4 players remain. Can you figure out how many matches will this stag have now?
hallow crane
then i have to do it for 3+2+1 (4 player stage)
this?
kindred pier
If $n$ players do one on one fight against every other player, then it is $\frac{n(n-1)}{2}$ total combinations.
solid kilnBOT
SWR
sleek aurora
this?
Yes
Now in the next stage, only 2 players remain, so only one match, the final
solid kilnBOT
Result:
35
I'll check and let you know
hallow crane
If $n$ players do one on one fight against every other player, then it is $\frac...
how?
wait
kindred pier
Any first player is n choices. Then n-1 other choices for second player. The divide by 2 for repetition.
solid kilnBOT
Tardis
kindred pier
$1+2+\cdots +(n-1)=\frac{n(n-1)}{2}$
The genius of math
hallow crane
$1+2+\cdots +(n-1)=\frac{n(n-1)}{2}$
oh yeah it's just an arithmetic sum
hallow crane
If $n$ players do one on one fight against every other player, then it is $\frac...
btw where did you get this formula from
I get what tardis has shown ofc
kindred pier
btw where did you get this formula from
logical conclusion
hallow crane
S = 1+2+3+...(n-1)
S = (n-1) + (n-2) + ...1
2S = (n)(n-1)
S = (n)(n-1)/2
sleek aurora
Any first player is n choices. Then n-1 other choices for second player. The div...
See this
hallow crane
uness you tried too
kindred pier
Degeneracy of pairs. (x, y) and (y, x) should be considered the same.
hallow crane
Degeneracy of pairs. (x, y) and (y, x) should be considered the same.
okay yeah i get that, but honestly a bit confused as to how you knew there'd be like exactly twice of those
kindred pier
Only two degenerates in pairs: (x, y) and (y, x)
hallow crane
,, \frac{{n \choose 1} \cdot {{n-1} \choose 1}}{2}
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斗地姿
hallow crane
Only two degenerates in pairs: (x, y) and (y, x)
Hmm
kindred pier
Or just n choose 2
hallow crane
I guess i'll think about it lol
hallow crane
Or just n choose 2
okay using your formula i'd have
ah okay i think i get it
the intuition is hard to get tbh lol
i can derive the closed form, but that's about it
hallow crane
Only two degenerates in pairs: (x, y) and (y, x)
okay i tried a smaller set:
{1,2,3} and yeah i see your point
half are basically the exact same pairs
i can generalize it for (x,y,z) and (x,y,u,t) or something like that as well right?
There'd be three degenerates in pairs for the former and 4 for the latter?
kindred pier
three degenerates?
hallow crane
if i'm working with
{1,2,3,4} and i wanted to choose 3 numbers
would it be (n)(n-1)(n-2)/3?
what that's not right
4C3 = 4
sleek aurora
{1,2,3,4} and i wanted to choose 3 numbers
Choosing 3 numbers is same as leaving out 1. So you can leave out one out of four in 4 ways
sleek aurora
would it be (n)(n-1)(n-2)/3?
Also, the number of degenerates won't be 3, but 6.
123, 132, 213, 231, 312, 321 all mean the same
hallow crane
Also, the number of degenerates won't be 3, but 6.
123, 132, 213, 231, 312, 32...
owhh
so the number of degenerates is basically n!?
sleek aurora
so the number of degenerates is basically n!?
That's correct
hallow crane
I see
I'm probably just going to use the combinations formula lol
Is there a video or something u can recommend to build this intuition? Ig this is literally just how combinations work but i guess i'm missing something
sleek aurora
Is there a video or something u can recommend to build this intuition? Ig this i...
I don't know about a video, but the intuition is fairly easy to build
Say you have to choose r from n. For first element, you have n choices, so you choose the first element. Now for the second element, you have n-1 choices, for third you have n-2 and so on. So for the rth element, you have n-r choices, hence the number of ways you can choose r elements (INCLUDING degenerates) is n(n-1)(n-2)...(n-r) which is essentially n!/(n-r)!
Now for r objects, there are r! degenerates, so you divide your thing by r!, thus finally giving n!/(n-r)!r! ways
hallow crane
@sleek aurora yeah that makes a lot of sense
i think i was pondering on your previous message that said r!(n-r)!/n!
i was confused why it was reversed
sleek aurora
Yeah I wrote it incorrectly
hallow crane
But yes i get your point, thanks @sleek aurora
sleek aurora
Sorry
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hallow crane
Say you have to choose r from n. For first element, you have n choices, so you c...
for permutations, i don't care about degenerates so i stop here right?
since i want to include degenerates
sleek aurora
That's correct. In permutations, ordering matters, so 123 and 132 are different
hallow crane
That's correct. In permutations, ordering matters, so 123 and 132 are different
also one more stupid question
let's assume there's 20 ways i can choose some elements (including degenerates)
and we'll assume like 5 degenerates
shouldn't it be 20 - 5?
Why are you dividing?
sleek aurora
5 degenerates for EACH choice of elements I am making
For example, say I have to choose 3 elements from {1,2,3,4,5}
sleek aurora
For first element, I chose 1, for 2nd I chose 2, for 3rd I chose 3. So my choice is 123. But the method I use to chose will consider 123 and 132 different, so I have to correct it. 123 gives 6 degenerates.
But what if I chose 234 instead?
That also gives me 6 degenerates.
hallow crane
ah so each "pair" (but extends to any n-element subset ig) has equal amounts of degenerates
okay fair enough
sleek aurora
For EVERY choice of three numbers I take (say 123), I will get 5 more choices which will be exactly same (like 231, 321, etc), for a total of 6 ways for getting ONE choice of three numbers. Hence we divide by 6
hallow crane
For EVERY choice of three numbers I take (say 123), I will get 5 more choices wh...
yep i get it
I guess that's all
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sonic stone
Linear Algebra help
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sonic stone
im stuck on the part where I have what I thought would work
ive been going up and down in the notes and honestly just need some direction
make sure to @ me when responding
gray rapids
for T to be linear that means T(u + v) = T(u) + T(v) if memory serves.
sleek aurora
3(x1+x2)-2(y1+y2) = 3x1+3x2-2y1-2y2 = (3x1-2y1) + (3x2-2y2)
And x1+x2 is, well, x1+x2
Do you see it now
@sonic stone
sleek aurora
Yes
sonic stone
ok im trying to find the second line where I entered 6 (that was just to fill it in)
because its just T(u) + T(v), would it just be (x1+x2)(y1+y2)?
gray rapids
well, the first box they want you to fill in they want a direct substitution to the transformation given if i understand right. simialr to what tardis is saying.
sleek aurora
because its just T(u) + T(v), would it just be (x1+x2)(y1+y2)?
No, that will only be x1+x2
sonic stone
ok ok
i get it now
i was putting x1 and x2 only into the top line and y1 and y2 into the bottom one
me putting wrong number places
otherwise I get it so thank you!
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scenic basin
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scenic basin
everything that’s blank i have no idea
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broken pilot
everything that’s blank i have no idea
for the first one, apply the power rules
,tex .log rules
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rysrobrgldvoelr👻ep>vneae=u
broken pilot
,tex .exp rules
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rysrobrgldvoelr👻ep>vneae=u
broken pilot
for the first question, the product rule
solar granite
is this where i get help
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solar granite
been on this for a while and cant find where i messed up i wrote every step that i did down
this is matrices
wraith hinge
is this your REF
solar granite
thats the final one yea
wraith hinge
been on this for a while and cant find where i messed up i wrote every step that...
too much to read, can you do the row operations again and carefully, document your operations with R2-> R1 + 3R2, swapped R1 with R2 for example
also no need to write down every single operation on each entry,
R2-> R1 + XR2 for example would be sufficient to plug into your calculator and get the entries
wraith hinge
too much to read, can you do the row operations again and carefully, document yo...
ignore, just redo the operations again
do the first operation and send to me i show you what i mean by documenting like RX->
wraith hinge
i think this is what ur asking for
yea much better
but you did not indicate the row that you're applying those row operations to
solar granite
oh okay
wraith hinge
and like i said the REF you got to is wrong, it would be much faster for the both of us and for practise sake for you to redo
solar granite
from the start?
wraith hinge
from the start?
yeah
solar granite
okay ima start from R1 <-> R2
wraith hinge
okay so starting with
$$\begin{pmatrix}
4 & -3 & 5 & 3 & 4\
1 & -1 & -1 & 1 & 6\
3 & 1 & -1 & 3 & 0\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
solar granite
i can do it on discord?
wraith hinge
embarassing tex
solid kilnBOT
okay so starting with
$$\begin{pmatrix}
4 & -3 & 5 & 3 & 4\\
1 & -1 & -1 & 1 & 6\\
3 & 1 & -1 & 3 & 0\\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
wraith hinge
so R1 <-> R2
any reasons to why
wraith hinge
okau so
$$\begin{pmatrix}
1 & -1 & -1 & 1 & 6\
4 & -3 & 5 & 3 & 4\
3 & 1 & -1 & 3 & 0\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
solid kilnBOT
wraith hinge
then -4R1+R2
to what row
solar granite
0 1 9 -1 -20
wraith hinge
$$\begin{pmatrix}
1 & -1 & -1 & 1 & 6\
0 & 1 & 9 & -1 & -20\
3 & 1 & -1 & 3 & 0\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
solid kilnBOT
$$\begin{pmatrix}
1 & -1 & -1 & 1 & 6\\
0 & 1 & 9 & -1 & -20\\
3 & 1 & -1 & 3 & 0\\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
solar granite
-3R1+R3
wraith hinge
,w -4(1,-1,-1,1,6) + (4, -3,5,3,4)
solid kilnBOT
wraith hinge
$$\begin{pmatrix}
1 & -1 & -1 & 1 & 6\
0 & 1 & 9 & -1 & -20\
0 & 4 & 2 & 0 & -12\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
solid kilnBOT
wraith hinge
,w -3(1,-1,-1,1,6) + (3,1,-1,3,0)
solid kilnBOT
wraith hinge
0 4 2 0 -12
uh check your calculation on the last entry
solar granite
okay give me a sec
fucking
HELL
WHY DID IT TAKE ME TILL NOW TO REALIZE
0 4 2 -6 -12
wraith hinge
what?
solar granite
what
wraith hinge
how are you getting -6
R3 = -3(1) + 3 ?
wraith hinge
oh the 3 is negative
deadly mistake on your original matrix
$$\begin{pmatrix}
4 & -3 & 5 & 3 & 4\
1 & -1 & -1 & 1 & 6\
3 & 1 & -1 & -3 & 0\
-1 & 2 & 2 & -1 & -7
\end{pmatrix}$$
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@solar granite Has your question been resolved?
wraith hinge
again you're not indicating where these operations are applied to
solar granite
they are applied to whatever is at the end
wraith hinge
okay
solar granite
like r1 +r4 is gonna be for r4
wraith hinge
this is what you should have got
wraith hinge
this is what you should have got
uh maybe not swapping that R2->R1 and just eliminating R2 with R2->R1+R2
wraith hinge
this is what you should have got
row 2 could've been zero (the first entry) by just adding it with row 4
1+ (-1)
wraith hinge
,w row reduce ((1,-1,-1, 1, 6), (0,1,9,-1,-20), (0,2,4,-6,-18), (0,1,1,0, -1))
solid kilnBOT
wraith hinge
okay now i found a way of checking errors
wraith hinge
P2 good checked, P4 bad
,w row reduce ((1,-1,-1, 1, 6), (0,1,9,-1,-20), (0,2,4,-6,-18), (-1,2,2,-1, -7))
solid kilnBOT
wraith hinge
,w (1,-1,-1,1,6) + (-1,2,2,-1, -7)
solid kilnBOT
wraith hinge
how did i fuck up that bad
wraith hinge
im used to it
wraith hinge
maybe i should quit computer science if this isnt even the start
uh
when i started i ddint get it in minutes
solar granite
well atleast i could put the hours in
atleast im not bored when doing it
i just hate it
cause of how long it is
well thanks bro thats it
wraith hinge
well thanks bro thats it
👍 no worries, close to free channel
solar granite
how u close
wraith hinge
.coose
solar granite
.close
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Available help channel!
quasi karma
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quasi karma
back again with formulae 💔
vernal warren
Make x be alone on 1 side
vernal warren
Ik
vernal warren
Faster way would be to factor/take common p
quasi karma
then q=p(x+3)?
vernal warren
Yeah
quasi karma
then i factorise?
vernal warren
q/p - 3 is not equal to q-3/p
vernal warren
No
vernal warren
$\frac{q}{p} - 3$
solid kilnBOT
JustToPro
quasi karma
ya
vernal warren
Yeah give spaces
vernal warren
Np
quasi karma
.close
quartz plume
here's my problem
the method i used was first by rewiting a = b (mod m) as:
-
a = mq + r, where q is some integer and r is the remainder
-
b = mk + r, where k is some integer
and then following this i rearranged each equation and got:
-
a - r = mq
-
b - r = mk
from this i deduced that :
- m | (a -r)
- m | (b - r)
which i then concluded that :
- a = r (mod m)
- b = r (mod m)
quartz plume
is my conclusion valid?
it isnt the final solution
but i just wanted to make sure that im allowed to perform those last few steps
specifically this:
from this i deduced that :
m | (a -r)
m | (b - r)
im not sure if this is a valid claim
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@quartz plume Has your question been resolved?
quartz plume
<@&286206848099549185>
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@quartz plume Has your question been resolved?
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@quartz plume Has your question been resolved?
covert oxide
I started it
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I started it
covert oxide
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@covert oxide Has your question been resolved?
covert oxide
<@&286206848099549185>
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@covert oxide Has your question been resolved?
covert oxide
<@&286206848099549185>
maiden hare
!status
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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
2
actally
1.5
i didquestion 7
so i have a smallchart
but idk whereto go fromthere
maiden hare
Show us
maiden hare
,rccw
solid kilnBOT
maiden hare
I find the question is a bit odd but whatever
You have a = -1 and you know the positions at t=0 and t=1
covert oxide
uhuh
maiden hare
You want the velocity at t=0
covert oxide
yes
maiden hare
I'm not sure what level of maths this is, do you know derivatives?
covert oxide
I'm not sure what level of maths this is, do you know derivatives?
im taking precalc rn
maiden hare
Ok let me think of something
covert oxide
ok
maiden hare
Ok so to go from 0m to 1.5m in 1s, how much speed do you need?
covert oxide
1.5m/s
maiden hare
Then to go from 1.5m to 2m in 1s?
covert oxide
5m/s
maiden hare
You mean 0.5?
covert oxide
yea
maiden hare
I don't quite understand what you mean
maiden hare
Remember position at t = 0 is 0, and at t = 1 is 1.5
covert oxide
then 1?
maiden hare
Surely at t = -1 it should be negative
covert oxide
but theres no negative in the multiple choices
maiden hare
No I'm asking about the position not the velocity
covert oxide
Right, so you see how velocity decreased by 1 in 1s
.
shount it incraease
so then
2.5?
maiden hare
That would be the speed you need between t = -1 and t = 0
but again, I'm asking about position
covert oxide
ok im confused
maiden hare
Hm maybe I shouldn't have brought up t = -1 haha
Let's do something else
We know that to go from 0m to 1.5m in 1s you need an average velocity of 1.5m/s
Then to go from 1.5m to 2m in 1s it's 0.5m/s
Then to go from 2m to 1.5m in 1s it's -0.5m/s
covert oxide
ok
maiden hare
So again, velocity decreases by one per second, linearly
So, say you know the velocity at t = 1 is v1 and at t = 3 is v3, what would be the velocity at t = 2?
covert oxide
v2
covert oxide
v*2
maiden hare
What's v
maiden hare
I mean yeah you can make it a function, v(1) = v1 and v(3) = v3, we want to find v(2)
maiden hare
Well you answered something else correctly but that's not v(2)
maiden hare
Wait no, when I write v1 or v2 or v3 it's not a multiplication
covert oxide
oh
maiden hare
You can pick a different letter if you want it doesn't matter
Like, you can say the velocity at t = 1 is p, and the velocity at t = 3 is q
p and q are just values you don't know
covert oxide
so how amisupposedto find at t=2
maiden hare
That's the point
You can use p and q to express the velocity at t = 2
It's still a value you don't know but at least you will have a relation
maiden hare
There is no variable, only three fixed unknown values we want to link
maiden hare
No we don't know that value
maiden hare
It's not
covert oxide
why
maiden hare
Have you worked with graphs before?
maiden hare
Can you try to graph the position here?
maiden hare
The x axis is time and the y axis is position
Just draw something and put some values in
covert oxide
maiden hare
Put some values on it for time and position
maiden hare
Ok now draw a vertical line from the 2 on the x axis until you meet the graph
covert oxide
maiden hare
Now at the intersection with the graph, draw a horizontal line (should meet the 2 on the y axis)
covert oxide
maiden hare
See how it just touches the graph without crossing it?
covert oxide
yea
maiden hare
That means the graph itself at that point is horizontal
The direction (slope) of the graph at any point is the velocity
covert oxide
ok
maiden hare
Of course that's just a drawing, you want actual calculations
covert oxide
😭
maiden hare
No, it asks for the velocity at t = 0, not t = 2
covert oxide
ok
maiden hare
Ok so because the acceleration is constant (-1), we know the velocity changes at a constant rate
covert oxide
yes
maiden hare
In this case it goes down by 1 every second
covert oxide
ok
maiden hare
So if it's, say, 46 at t = 1, what would it be at t = 2?
covert oxide
45
maiden hare
Right, if the position at t = 1 and at t = 3 is the same (we don't care what exactly), what would be the average velocity between t = 1 and t = 3?
maiden hare
Of average velocity?
covert oxide
i mean
since they are the same
the average is just t=1 and t=3 whichever one right
maiden hare
No, the positions are the same, that's different
covert oxide
Let's do something else
We know that to go from 0m to 1.5m in 1s you need an *av...
but didnt u say this
maiden hare
Yeah, 0m and 1.5m are not the same positions
covert oxide
it travels from 0m to 1.5 m in 1 second
so 1.5/1 = 1.5
so isnt 1.5m/s our average velocity
maiden hare
But the position at t =1 is not 0m and the position at t = 3 is not 1.5m
covert oxide
But the position at t =1 is not 0m and the position at t = 3 is not 1.5m
is t = time
maiden hare
Yes
covert oxide
at t=3 its 1.5m
maiden hare
Oh right my mistake, but still at t = 1 it's not 0m
covert oxide
yea it is
maiden hare
No?
covert oxide
maiden hare
maiden hare
Yeah so it's not 0m...
maiden hare
You are confusing me lol
covert oxide
idk
covert oxide
yes
covert oxide
yea
maiden hare
So if the position is 1.5m at both times, why would the velocity be 1.5m/s?
maiden hare
1.5m traveled in 1 second is what happens between t = 0 and t = 1
covert oxide
yea
maiden hare
Not between t = 1 and t = 3
maiden hare
If you start at your house at 2pm and go somewhere 3km away (the park or whatever) and you stay there between 3pm and 5pm and then you return home at 6pm
What's your velocity between 3pm and 5pm
maiden hare
Right
So do you agree that in the problem the 1.5m is just a position
It's just "1.5m away"
covert oxide
uhhuh
maiden hare
So if the car starts there at t = 1 and finishes at the same place at t = 3, what's its average velocity?
covert oxide
0
maiden hare
Right
maiden hare
Remember velocity can be negative
maiden hare
It's positive between t = 1 and t = 2 but then it's negative between t = 2 and t = 3