assuming 4+ is also good
#help-38
shrewd ridge
rugged latch
azure zenith
Need help with this, I ended up with $$\textrm{Volume}=\int_{1}^{2}\int_{0}^{2\pi} (r\cos t+r\sin t)rdtdr$$
solid kilnBOT
Ericsson
uneven eagle
you would need a triple integral here, no?
azure zenith
Yeah but the bounds of z are $0<z<x+y$ which simplfies to the above integral
solid kilnBOT
Ericsson
uneven eagle
ahh yes i see
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@azure zenith Has your question been resolved?
cloud geyser
I fail to see how x + y = z is a paraboloid
that looks like a plane to me
I'd say it's either a typo or it wants you to find the volume under the z = x + y plane bounded by the two cylinders and xOy 🤔 I might be mistaken though
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@azure zenith Has your question been resolved?
azure zenith
I fail to see how x + y = z is a paraboloid
I now see the problem here, the equation should be z=x^2+y^2 instead of z=x+y
That's funny
azure zenith
I'd say it's either a typo or it wants you to find the volume under the z = x + ...
How will you do that, say that instead of a paraboloid i had a plane z=x+y
cloud geyser
i think youd have to use triple integrals for that
trying to use x+y written in polar coordinates indeed results in 0
azure zenith
$$\textrm{Volume}=\int_{1}^{2}\int_{0}^{2\pi} \int_0^{rcost+rsint}rdzdtdr$$
solid kilnBOT
Ericsson
azure zenith
That should be it
cloud geyser
yeah
azure zenith
Hard to imagine that the volume is zero
cloud geyser
since we know for sure that z goes from the xy plane up to the maximum point of x + y (within the constraints), that triple integral should work
cloud geyser
I now see the problem here, the equation should be z=x^2+y^2 instead of z=x+y
do you know how to move forward from here? 🙂
azure zenith
Yup
cloud geyser
nice!
misty raptor
can someone explain this step to me
misty raptor
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vapid shore
Can someone please help me
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Can someone please help me
vapid shore
How did use LCT
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wise pecan
help?
daring dragon
Is this right?
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Is this right?
@daring dragon Has your question been resolved?
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sharp owl
why is (n-1) + n?
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why is (n-1) + n?
sharp owl
nimble stone
its a sum of terms,
n-1 is the number before n, n-2 is the number before that etc
n is the last number
if n was 6 for example
the sum would just be 3+4+5+6 5 is n-1, 4 is n-2 etc
sharp owl
if n is 1 = 1-1 + 1 = 1?
sharp owl
n must be greater than or equal to 3
where i can practice sigma notation
nimble stone
search summation questions or sigma notation questions, youll find some somewhere
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open fulcrum
help
open fulcrum
wraith hinge
How'd you get that answer?
open fulcrum
it asks to simplify and when inputting 2n sqrt2 it returns incorrect
open fulcrum
How'd you get that answer?
4n^2 + 4n^2 = c^2
open fulcrum
ohhhhh
wraith hinge
@open fulcrum Well, is that it? If so, don't forget to .close
open fulcrum
.close
opal goblet
So for this simultaneous equation, I’m confused what the solution is supposed to be
opal goblet
would you say that there is only no solution if x=3y?
fair patio
well yes there is no solution if x = y or x = 3y
because if x = 3y, you'd end up with this
$9y^2 - 45y^2 + 6y^2 = 2$
solid kilnBOT
Charles of Palestine
solid kilnBOT
Charles of Palestine
fair patio
therefore y would be equal to $\sqrt{\frac{2}{-30}}$
solid kilnBOT
Charles of Palestine
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opal goblet
alr that makes sense but how come x=y is also no solution?
fair patio
because no value multiplied by 0 will ever equal two
so no solution exists for either case
fair patio
So for this simultaneous equation, I’m confused what the solution is supposed to...
the second to last equation on the right,
$9y^2 - 15y^2 + 6y^2$, 9 + 6 - 15 = 0
so $ 0y^2 = 2$
any number times 0 equals 0, and 0 != 2
solid kilnBOT
Charles of Palestine
fair patio
this != means not equal btw
opal goblet
so if you were the plug in the answers on the left
x=+-1
y=+-1
would it not make the simultaneous equations true?
fair patio
that is true for x = y
whether x is positive 1 or negative 1, 0 * 1 = 0
and 0 != 2
0 * -1 = 0
opal goblet
sorry, so the simultaneous equation would only be true if both x=y and x=3y?
warm kernel
sorry, so the simultaneous equation would only be true if both x=y and x=3y?
it's only true if x = y, there is no solution for x = 3y according to the solution you gave
ionic wing
i need help with this optimization problem
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i need help with this optimization problem
ionic wing
i'm not sure why my p(x) equation is wrong, i'm pretty sure i've done everything right to get it
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fathom ivy
have you tried substituting in your value and make sure it matches the problem description
ionic wing
have you tried substituting in your value and make sure it matches the problem d...
i made the wrong value my x it was really dumb i got it
.close
lone mauve
is this correct?
trim joltBOT
is this correct?
trim joltBOT
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Available help channel!
rain spire
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latent rover
4 red balls, 1 blue, 1 black, if we took 2 balls simultaneously, what is the probability that among the 2 balls, 1 is black
latent rover
1C1 * 5C1 / 6C2?
shrewd ridge
makes sense
latent rover
I think you are right
Number of wanted outcome/all outcomes
i think that i'm right too, now we two think that i'm right
"think"
we need another vote
latent rover
Why are you doubting?
our exam
latent rover
Cant you just run simulation writen on programing language
simultaneously, means both at once?
latent rover
Cant you just run simulation writen on programing language
how
stone kiln
simultaneously, means both at once?
There is no different. Probability be the same
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earnest breach
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earnest breach
Did they get b_n through fraction decomposition ?
err maybe not that cause it would have two fractions.
could I do that though if I wanted?
zinc ginkgo
they're just comparing the highest powers of the numerator and denominator
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earnest breach
they're just comparing the highest powers of the numerator and denominator
remeind me, if I rewrote root(n+2) as (n+2)^1/2, would the power be 1/2?
solid kilnBOT
riemann
sand obsidian
What steps do I take to get the absolutvalue of $$\frac{3+i}{4+3i}$$ ? do I start with conjugating denominator?
solid kilnBOT
Totalani
zinc ginkgo
Yes
sand obsidian
ok let me do that
$$\frac{15-5i}{25}$$ this is what I get, this is now where I get kinda stuck
solid kilnBOT
Totalani
sand obsidian
oh wait
its absolutvalue
so the i isnt suppose to be there
so its 10/25
or $$\frac{\sqrt{ 10 }}{5}$$
solid kilnBOT
Totalani
sand obsidian
am I thinking correct?
lyric nymph
so consider the complex number z=a+bi then |z|=sqrt(a^2+b^2) right ?
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lyric nymph
np you did everything
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molten comet
Hello
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molten comet
.close
wraith hinge
<siunitx, tikz>[
\DeclareSIUnit{\nc}{\newton\per\coulomb}
\DeclareSIUnit{\g}{g}]
\textbf{Question:} What must the charge (sign and magnitude) of 1.46\si\g\; particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 680\si\nc?
\vs{3 mm}
\textbf{My confusion:}
So like i tried drawing a free body diagram to get a better idea but i am like getting this
\env{center}{
\env{tikzpicture}{
\draw (0,0) circle (0.5) node[below] {};
\draw[->, thick] (-0.3,-0.4) -- (-0.3,-2) node[below] {$w$};
\draw[->, thick] (0.3,-0.4) -- (0.3,-2) node[below] {$E$};
}
}
Which doesnt quite make sense lmao so what is going wrong
stoic iron
Well you would choose the sign of charge of the particle so that E points upwards
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uneven steeple
<:thonk:406575732563902485>
so for a negatively charged particle, it would go in the opposite direction
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lone mauve
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lone mauve
k = -4t + 3x?
no clue if that's the correct idea here..
consistent = one or infinite solutions, right?
inconsistent = no solution
but do I call -3 "x", and do I call -4 "y"?
y = t because it's a free variable
it's not specified in the matrix format, they are just numbers
also, are these augmented matrices?
i don't see a line to the left of the last column
maiden hare
The line is optional
lone mauve
should be mandatory
maiden hare
They are just matrices
If you use a matrix for a system of equation like in here, then it's an augmented matrix
maiden hare
Give me a minute
Hm maybe I'm not the best person to help here but to me it looks like k has only one possible value
lone mauve
what would you put for k instead?
my answer for b is not much better..
i believe the row operations are allowed, but I don't know if I made things even more complicated by doing this
lone mauve
and -3x = -5/2 + 4?
lone mauve
but I don't need to find for x or y, right? just k
this is all just extra work for no reason
maiden hare
Why -15/6 ??
lone mauve
oops that should say 5/6
maiden hare
Ok yeah I guess you only need k
lone mauve
alright
i'm going overkill
so for b)
i guess same idea, going overkill
I just need k
lone mauve
maiden hare
How did the -kR1 + R2 -> R2 operation add some k's in the first row?
lone mauve
ooops
maiden hare
And this should be -1/2 k - 2
maiden hare
Not really tbh
maiden hare
I mean I don't know what would be the proper way to solve this but just by looking at the matrix you know k = -4
lone mauve
are k and 4 both on the "x" column?
maiden hare
Yeah
maiden hare
kx
This reads as kx + y = -2 and 4x - y = 2
Add the two equations, you get (k+4)x = 0
lone mauve
This reads as kx + y = -2 and 4x - y = 2
maiden hare
Lol that's one way to go about it
I just think you should take a moment to contemplate the fact that the thing you circled simplifies to -4
lone mauve
hmmm
lone mauve
I just think you should take a moment to contemplate the fact that the thing you...
how did you get 4 for k?
I found x and y if that helps
but they are dependent on each other
maiden hare
Add the two equations, you get (k+4)x = 0
.
maiden hare
how did you get 4 for k?
-4, and do you mean from the problem or from that thing you circled?
lone mauve
so the top row is irrelevant
if I see 0's on a row, that tells me something
kx = -2 -y is not relevant here to solve for k?
like could I also write this for the answer?
maiden hare
It is, you used it with the second row to make x(k+4) = 0
maiden hare
like could I also write this for the answer?
I mean these are true statements
lone mauve
so while this is true, it's actually the wrong direction
I want to remove x and y from the answer for k=
maiden hare
But if you can get k = a concrete value, then that's called a solution for k
tulip violet
like could I also write this for the answer?
All you did was rewrite the top row as an equation equal to k, there was no point in doing this
lone mauve
q = 2x + 4 is not a solution for q?
maiden hare
No
tulip violet
Your best option is to row reduce the matrix
lone mauve
OK
maiden hare
q = 2x + 4 is not a solution for q?
Well, depends on the problem
lone mauve
I think the top row k is messing me up
I wanna solve for k there from the first row because it's already on it's own
but it's actually kx, not k
tulip violet
And see what value of k would make it consistent and inconsistent
lone mauve
alright that makes sense
so we are finding where x = 0 to solve for k
to allow for 0 + 0 = 0 on row 2, which is the only true statement available for that row
so this system has infinite solutions
and k = -4
is that correct? or is it only one solution
if it was an entire row of 0's it would be infinite solutions
maiden hare
It's infinite solutions of the form 4x-y = 2
lone mauve
so in other words, this augmented matrix (system of equations) has infinite solutions
maiden hare
Yes if k = -4, its solution space is a line
lone mauve
I don't get it though..
-4 = -2 + y
y = -2
maiden hare
If k =/= -4 then it has no solution
maiden hare
No it's -4x = -2 + y ...
lone mauve
oh right
so we will never know what x and y are?
in terms of constant numbers
only k?
tulip violet
No you can still find what x and y are
x and y represent the solution vector
You just have different cases on what k can be to make the solution consistent or not
maiden hare
No you can still find what x and y are
Not as constants
tulip violet
If you row reduced that matrix, you'll see this
Where you can determine that what k is
tulip violet
Not as constants
Not as a unique solution, you'll have infinitely many
maiden hare
2?
tulip violet
this gives us 2 solutions for variables right here
There is not a unique x and y solution
tulip violet
There are infinitely many
lone mauve
x= 0
y = -2
k = -4
maiden hare
That gives you nothing about x and y
lone mauve
what are the others?
tulip violet
If you row reduced that matrix, you'll see this
Where you can determine that wh...
Because if you do this, k = -4, you have a row of zeros therefore y is a free variable, then you do the parametric solution form
maiden hare
Told you, it's of the form 4x-y = 2
lone mauve
oh right
lone mauve
Told you, it's of the form 4x-y = 2
both forms
so what are other solutions?
x= 0 y = -2 k = -4
k can be anything is that the idea??
lol sorry, still confused..
if I plug these values in
that's how I read it
tulip violet
You are focused too much on x and y
lone mauve
OK..
maiden hare
If k is anything other than -4, you get no solution
lone mauve
only because I found them here
maiden hare
If k = -4, you get a line 4x-y = 2
That means you can pick whatever value for x and deduce y, or vice versa
tulip violet
If k is anything other than -4, you get no solution
That's actually wrong, if k is not 4, you get a unique sol
maiden hare
That's actually wrong, if k is not 4, you get a unique sol
How?
lone mauve
well anyways, it's only asking for k here
I think I found it
but it's asking for "all values of k"
is there only one value of k?
maiden hare
Oh yeah you're right you get x = 0 and y = -2
maiden hare
You forgot a y
solid kilnBOT
tulip violet
There it is
lone mauve
why 3x?
lone mauve
wait so k has infinitely many solutions?
tulip violet
If you row reduced that matrix, you'll see this
Where you can determine that wh...
Going back to this
lone mauve
not just -4?
tulip violet
You only have two cases for k, when k = -4 and when it doesn not equal -4
If k = -4, you get infinitely many solutions
If $k \neq -4$, you get a unique solution
solid kilnBOT
CaptainNova22
tulip violet
Those are the only two cases
tulip violet
You are overthinking it
It's asking for the value of k
If k = -4, the system has infinite solutions
maiden hare
but I thought a matrix can only have one of the 3?
That's still true, but you're not dealing with one matrix, you're dealing with an infinite number of different matrices based on the value of k
tulip violet
If k does not equal -4, you have a unique solution
lone mauve
so when you see a variable in a matrix like this (k), does it indicate there will be (infinite && one solution) || (no solution)?
tulip violet
That's still true, but you're not dealing with one matrix, you're dealing with a...
Technically I would say two matrices, one for k = -4 and k not equal to -4 because for k not equal to -4, when you reduce it, you'll get the same solution no matter the k
lone mauve
but never (infinite) || (one solution) || (no solution)?
when you see a variable in a matrix
lone mauve
but never `(infinite) || (one solution) || (no solution)`?
or can it sometimes be this too?
tulip violet
when you see a variable in a matrix
It just depends on the system
If you have
-4x + 1 = -2
4x + h = k
There are two variables, and multiple situations
maiden hare
or can it sometimes be this too?
You can get any combination
tulip violet
Because if you row reduce it, you get this
So if h = -1, and k does not equal 2 then you get no solution
But if h = -1 and k = 2, then it's infinite, but if h doesn't equal -1, then you have a unique solution
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zealous fossil
What is 5+5?
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What is 5+5?
wraith hinge
What is 5+5?
use a calculator
knotty locust
<@&268886789983436800> Troll
nimble stone
.close
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merry gyro
What is 5+5?
Don't troll in help channels!
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Available help channel!
lone mauve
Because if you row reduce it, you get this
So if h = -1, and k does not equal 2...
sorry, back now - but doesn't it also depend on what y is?
lone mauve
like where do you go from this point to determine how many solutions there are?
what I have done here helps? or not really
0,0,0 = infinite solutions
0,1,2 = 1 solution
but we have variables, so we are actually setting the entire thing equal to 1 (any number) and to 0 (all 0's), right?
also, these are the same thing?
would it be (y + h) or (y + hy)?
and it's only row 2 that I set to 0, and to 1 (or any number)? to check for solutions
infinite, unique, and none
lone mauve
just an example
lets look at this one again
k = -4 gives infinite solutions
i think
because 0,0,0 for row 2
austere cedar
That's right
austere cedar
Any other
lone mauve
lone mauve
*Any other*
no that doesn't make sense tho..
let's look at row 2...
5+1 for example
if k=5
6 + 0 = 0?
that is not a solution
austere cedar
Like, let's say k = 2. Then that bottom equation would read,
6x = 0
But there's a unique solution to that, x = 0
lone mauve
hmmm
so we are saying x=0?
wait so what is the unique solution? if x=0 it would also be infinite solutions
austere cedar
In fact, if we are assuming k ≠ -4, let's divide row 2 by k+4
We have:
|k 1 | -2|
|1 0 | 0|
x = 0 pops out
lone mauve
huh.. and xk = -2 -y?
austere cedar
Well, we're not done. We can keep reducing
austere cedar
In fact, I'll swap the rows, then multiply the top by k:
|k 0 | 0|
|k 1 | -2|
Subtract the top from the bottom and put that back in:
|k 0 | 0|
|0 1 | -2|
To get y = -2
(0,-2) is always a solution, independent of k
If you multiply that by the matrix, it's pretty clear how this works. Letting x = 0 causes k to go away
lone mauve
In fact, if we are assuming k ≠ -4, let's divide row 2 by k+4
are we allowed to use variables for scalars with row operations?
austere cedar
k is better thought of as a constant. We can choose it
If k = 2, then I'm just dividing that row by 6. Nothing crazy about that
lone mauve
k is better thought of as a constant. We can choose it
so all of this is perfectly legal?
now we found value for y, but how does that help us?
the question is "find all values of k"
lone mauve
so this is where it can can dicey?
austere cedar
Yep exactly. If you are dividing by a constant we can change, it's worth mentioning what that constant can't be
lone mauve
k = -4 gives infinite solutions
but i thought we said k = -4 gives infinite solutions? this is not true?
austere cedar
So note your work gives us the unique solution (0,-2)
If k = -4, we can't actually complete your work
And need to assess the matrix some other way
You've already done that way back, and found that k = -4 gives infinite solutions
lone mauve
so even though we found k = -4 makes it 0, it is out of the domain to begin with
so the only value that gives infinite solutions is when x=0?
austere cedar
We can't choose x, only k
lone mauve
yeah
lone mauve
why (0,-2)?
k = -4 gives infinite solutions
but k = -4 is also out of domain for multiplication of 1/(k+4)
so I'm a bit confused
austere cedar
If you fully reduce your matrix (which you can now finish off by dividing the top row by k) you get:
|1 0 | 0|
|0 1 | -2|
That's the unique solution
lone mauve
yes
austere cedar
THIS REDUCTION WAS ONLY POSSIBLE IF K ≠ -4
lone mauve
so x = 0
y = -2
OH
(0,-2)
so k=-4 gives infinite solutions
k = -2 gives on unique solution
austere cedar
The second k becomes -4, you won't be able to reduce this to a unique solution
lone mauve
do we mention anywhere with this, that k cannot be -4?
k = -2 gives on unique solution
for the final answer
austere cedar
k = -2?
lone mauve
I finally understand now
austere cedar
(x, y) = (0,-2)
But only for the case where k ≠ -4
lone mauve
k cannot be -4 for unique solution
but it must be -4 for infinitely many solutions lol
austere cedar
If k = -4, we get infinite solutions
lone mauve
that's confusing af lol, i think I just gotta get used to that with systems of equations
austere cedar
Imo try a few more reductions, and note when you're dividing by something that could be 0
lone mauve
(x, y) = (0,-2)
But only for the case where k ≠ -4
do we need to mention domain restriction, or no need?
when answering this
just this is fine?
k = -4, infinite solutions k = -2, unique solution
austere cedar
There's no domain restrictions. k is allowed to be -4, but if you plug that in, your matrix won't reduce down
austere cedar
No. I don't see where you got k = -2
lone mauve
There's no domain restrictions. k is allowed to be -4, but if you plug that in, ...
hmmm
oh
i'm confusing y = -2 for k = -2
lemme think about this some more
austere cedar
Sure, hit me up if you have a question
lone mauve
ty
it's pretty clear that k can also be 0 here
to give infinitely many solutions
so k= -4 and k= 0
austere cedar
Actually, could be helpful,
Take your original matrix, multiply it by (0,-2). Show that it does not matter what k is, this always solves the system
lone mauve
oh right it's actually kx
not just k
but if k = 22 for example
that wouldn't work.. or would it?
22x = 0
x is still 0?
22x = 0
x = 0/22
x = 0
interesting...
so k is all real numbers
except for -4
austere cedar
I actually meant like the ORIGINAL matrix. That is,
|k 1| | 0|
|4 -1| times |-2|
lone mauve
final answer for infinitely many solutions?
for finding a single unique solution i'm still not sure
I only see infinite solutions with this augmented matrix
even after accounting for that one domain issue, it's still infinite
austere cedar
I actually meant like the ORIGINAL matrix. That is,
|k 1| | 0|
|4 ...
I want you to actually try this matrix multiplication
lone mauve
Showing (0,-2) is always a solution, independent of k
hmmm
i have to think about that..
austere cedar
Because multiplying your matrix by (0,-2) gets the augmented part
lone mauve
Because multiplying your matrix by (0,-2) gets the augmented part
the solutions?
the last column of augmented matrix always = solutions, right?
or maybe a better word would be: constant?
the other columns are variables
but also solution, for last column
because the line down the left side of the last column on an augmented matrix is like an equals sign
austere cedar
You are solving the matrix equation
Ax = b
A is the actual body of the matrix, left of that line
b is "the augmented part", right of that line
x is the thing you're solving for, the "solution".
The question is "what do we multiply the matrix by, to get the augmented part"?
You've just shown that (0,-2) is a solution. You multiplied it by your matrix, and the augmented part was the result
Sometimes we refer to this in a system of equations form, where x = 0, y = -2. Same idea.
lone mauve
interesting..
lone mauve
You are solving the matrix equation
Ax = b
A is the actual body of the matrix, ...
could I not just eliminate k entirely?
austere cedar
If you can divide by 4 + k, then you can indeed eliminate k entirely
austere cedar
But that's not possible if k = -4
Actually, another interesting exercise. What happens if k = -4? Let's put that into the matrix equation:
|-4 1 | -2|
|4 -1 | 2|
lone mauve
If I add them together I get 0 0 0
austere cedar
And if we try reducing:
|-4 1 | -2|
|0 0 | 0|
I cannot reduce completely anymore
We're left with the line
-4x + y = -2
And that's infinitely many solutions
.
If k = -4, there's infinitely many solutions, because the matrix cannot be fully reduced.
If k is anything else, the matrix can be fully reduced. There's a unique solution (0,-2).
lone mauve
OK
ty for the help @austere cedar
saved a screenshot of this, I will have to review again tomorrow
appreciate the help
austere cedar
Hopefully it made sense haha. Feel free to ask if you have anything else
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languid bluff
Yall so question:
I got this nasty lookin elliptic integral function that im trying to approximate into a maclaurin series (i have to integrate it another time so thats why). Does anybody think they could help me calculate that out and write it to the term where the remainder is negligible? Provided is the original elliptic integral (the elliptic integral is of the second kind), as well as the first derivative of it
languid bluff
Ignore the knees its midnight and im too tired to crop 💀
Im really sorry ik it hasnt been 15 minutes yet but i really gotta sleep soon and get this done so <@&286206848099549185>
Its not too hard of a problem i just need help with calculating a maclaurin series
😔
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rose grotto
Yall so question:
I got this nasty lookin elliptic integral function that im try...
nice legs 🤤
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wraith hinge
<tikz;siunitx>
\textbf{Question:} A \SI{+8.75}{\mu C} point charge is glued down on a horizontal frictionless table. It is tied to \SI{-6.50}{\mu C} point charge by a light, nonconducting \SI{2.50}{cm} wire. A uniform electric field of magnitude \SI{1.85e8}{N/C} is directed parallel to the wire. What is the tension in the wire?
\vs{3 mm}
My illustration of the problem is as follows:
\env{center}{
\env{tikzpicture}{
\node (a) [draw, shading=ball, ball color=b!50!bg, circle, inner sep=2pt, "\SI{-6.50}{\mu C}" below] {$-$};
\node (b) [draw, shading=ball, ball color=r!50!bg, circle, inner sep=2pt, "\SI{8.75}{\mu C}" below, right=3 of a] {$+$};
\draw [thick, gray] (a) -- (b);
\draw [|<->|] ([yshift=.5cm] a.east) -- node [fill=bg] {\SI{2.50}{cm}} ([yshift=.5cm] b.west);
\draw [->, shorten >=1cm, shorten <=1cm, ultra thick, purple] ([yshift=1.5cm] a.east) -- node [above] {$\vec{\vb* E}$} ([yshift=1.5cm] b.west);
}
}
wraith hinge
Anyways I'm not sure how to consider the freebody diagram here. How can I draw this out for the tension?
wraith hinge
Elementary physics questions are allowed here as per the moderators' decision.
atomic socket
idk physiscs im bad sorry
wraith hinge
like what counts as tension here? I'm super confused because aren't there only like the electric forces and the negligible weight?
.close
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weak ridge
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languid bluff
nice legs 🤤
Bro
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rose grotto
try thinking for each number which numbers relate to it
for example for [1], 1+b^2 being even means b^2 is odd so b is odd
but because of transivity that means [1] = {all odd numbers}
seems right
but the reasoning is bad imo
rose grotto
but because of transivity that means [1] = {all odd numbers}
.
is how i would write it
clever slate
also can i just use [1] and [2] as the equivalence classes when it could be like any odd or even number
rose grotto
and then after i know [1] id check [2] and with the same reasoning of 4+b^2 being even --> b is even --> [2] = {all evens}
and then i know im done
cause [1] U [2] = N
if you want to expand on the "because of transitivity" you can
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lone mauve
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lone mauve
just making sure, with dot product
this is true?
even with k(u . v . w) I could scramble them all around in any combination too?
n-tuples of equal size are 100% commutative with dot product, addition, and subtraction, and scalar multiplication
always
austere cedar
Yes that's all true
maiden hare
n-tuples of equal size are 100% commutative with dot product, addition, and subt...
n-tuples of equal size is a bit redundant
lone mauve
tyvm
so there is another term call cross product of n-tuples, right?
where instead of .
it's ×
u × v
maiden hare
Cross product is only for 3D vectors
lone mauve
oh, so the n-tuple must be of length 3 to use cross product?
u = (1,2,3,4)
v = (5,6,7,8)
u×v cannot work?
only u.v?
maiden hare
Yes
maiden hare
And if you're curious there's something called the exterior product or wedge product that kind of generalizes the cross product
AFAIK it's not very famous though, and people just use the cross product for everything 3D
maiden hare
An operator
maiden hare
XOR is sometimes written that way yes
full dock
ya
lone mauve
fascinating.. math merging into programming here
or maybe the other way around actually..
|| is usually or where it's one or the other, and it can also be both, it just cares if one evaluates to true
but XOR is one OR the other i guess, it's exclusive
maiden hare
I mean... computer science is basically maths wearing a costume
maiden hare
?
lone mauve
I mean... computer science is basically maths wearing a costume
it can involve, or is relatable to math, but I'm happy my BCompSci classes are not pure math
object oriented programming, compilers, syntax errors, are not really subjects that can be taught from pure math class
memory resource, if statements, loops, arrays, etc.
maiden hare
Ok theoretical computer science is maths wearing a costume
full dock
What’s the most bizarre maths branch that is borderline not really math but still in the realm of maths
lone mauve
actually only 3 pure math courses in the degree program
but more can be taken as electives if desired
lone mauve
MATH 204 is this program I am taking now (Matrices)
full dock
multi variable calc when
full dock
So like you’ve done triple integrals and stuff like that?
lone mauve
is that the disk washer method?
this looks different from that
I don't think we will be learning that in MATH 205
yeah that looks like Calculus 3
full dock
is that the disk washer method?
Disk washer is calculus 2
full dock
What the fuck
I just see the integral of sex show up and dissapear as an embed
have you done differential eqss
jovial tinsel
pls help
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final sage
Hint: only one of the lines 2) and 4) passes through (6,8)
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edgy jewel
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wraith hinge
how come this has no y intercepts
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how come this has no y intercepts
wraith hinge
even though it simplifies to $\dfrac{(x-1)(x-2)}{3(x-3)}$
solid kilnBOT
RecRio
wintry stag
evalutate it a 0
spare orchid
It simplifies to that for nonzero x
wintry stag
or, try to
spare orchid
you can't cancel two things on numerator and denominator if they're zero since you might be dividing by zeor
wraith hinge
subbing in 0 for x would give you 2 over -9
wraith hinge
you can't cancel two things on numerator and denominator if they're zero since y...
what does this mean
wintry stag
how come this has no y intercepts
when x = 0, what is this?
wraith hinge
0
red loom
not quite
wraith hinge
but it simplifies to something where it wouldn't be 0, which is my confusion
red loom
undefined?
for simplicity's sake yes 
we get division by zero
there is some more going on behind the scenes here
and you'll get to that in calc
but for now it seems you just need to know that 0/0
wraith hinge
I'm not taking calc but we've touched upon this in limits
red loom
essentially if we find the value really close to x=0 on the positive and negative
and if they match
then the limit exists
however $\lim_{x\to a}f(x)\neq f(a)$
wintry stag
We have that $\frac{x^3-3x^2+2x}{3x^2-9x} = \frac{x}{x} \cdot \frac{x^2-3x+2}{3x-9}.$ Notice that $$\frac{x}{x} = \begin{cases} 1 & x\not =0\ \text{undefined} & x=0.\end{cases}$$
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PajamaMamaLlama
wintry stag
yea i made a typo, oopsie
wraith hinge
maybe that's my mistake?
solid kilnBOT
Zander
wraith hinge
ok great, from there we can cancel the x's, correct?
wraith hinge
how come I get this when I simplify $\dfrac{(x-1)(x-2)}{3(x-3)}$
solid kilnBOT
RecRio
wraith hinge
is this wrong?
red loom
it's right only if x does not equal 0
wraith hinge
let's bring it back a bit
so
wait this makes no sense lol
so if at the start it's a invalid function
it'll always be an invalid function?
red loom
❌ it will only be invalid at x=0
wraith hinge
$\dfrac{x(x-1)(x-2)}{3x(x-3)}$
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RecRio
wraith hinge
so do the x's infront of the brackets in the num and den not cancel?
red loom
they only cancel if x≠0
wraith hinge
but we don't know x until we sub in for x
so your saying
we can't find a y-intercept with this equation
but with the initial equation we can
red loom
that's not really what I'm saying 
red loom
🎉
wraith hinge
but x can be zero in the simplified equation
wraith hinge
and is that not a valid step?
wraith hinge
so the algebra is dependant on what I sub in for x?
austere briar
No x = 7
wraith hinge
I don't get it lol
red loom
so the algebra is dependant on what I sub in for x?
in some specific cases, yes!
because we cannot divide by 0
do you remember what is means to actually cancel?
$\frac{2}{4}=\frac{1}{2}$ because $\frac{2}{4}=\frac{2\cdot\frac{1}{2}}{4\cdot\frac{1}{2}}=\frac{\frac{2}{2}}{\frac{4}{2}}=\frac{1}{2}$
wraith hinge
do you remember what is means to actually cancel?
to subtract?
or add
to remove terms
on opposite sides
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PajamaMamaLlama
red loom
the reason was can reduce fractions
it because we can divide on both sides (top and bpttom)
same thing for x
just as we did for 2
so if we divide out an x
the case where this results in division by zero is when x=0
solid kilnBOT
RecRio
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PajamaMamaLlama
red loom
notice how, normally, this would cancel the x's
but if x=0
then we get division by 0
which is hella cringe
wraith hinge
let me clarify
0 is because we're looking for a y-intercept
or there is literally a 0 in this function
wraith hinge
ok
wraith hinge
yeah that makes sense then
so x* 1/x normally equals 1
but if x is set as
0
0*1/0
is und
red loom
thank you for actually taking time to try to listen and learn
hope you have a wonderful evening
whole coral
Does that say that both K and L only depend on t? (Looks a bit cut off)
smoky kite
Oh yeah it says K and L depend only on t
whole coral
If so, then yea that looks fine to me, chain rule 
smoky kite
Instead of dQ on the top everywhere it wouldn’t be dF?
whole coral
Well pretty much the same thing (like if you said y = f(x) you can write either dy/dx or df/dx)
smoky kite
So either is fine?
whole coral
Yea should be! Though might be one notation is preferred to the other
smoky kite
In lecture and handouts it doesn’t say this version, so is it just to best to go by lecture etc
whole coral
In lecture and handouts it doesn’t say this version, so is it just to best to go...
Yeah better for consistency to match what they do, but at least you won't be surprised if you see others doing differently!
smoky kite
Oh ok tysm
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real herald
Can someone explain how to solve this lapse transform?
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Can someone explain how to solve this lapse transform?
real herald
I know its an unit funtion, but the table I am using from pauls notes does not have this form:https://tutorial.math.lamar.edu/classes/de/laplace_table.aspx
unreal loom
Can someone explain how to solve this lapse transform?
remember the time shift rule of laplace transforms
real herald
Not really, let me go back and review
ok im back @unreal loom
I looked over it and im still a little lost
is there any way you could let me know which funtion from the table to use?
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topaz oriole
Hello everyone!
having trouble with this math problem, chat GPT cant make sense of it apparently. I tried on my own
(X = $60,000,000 / 1.131)
topaz oriole
I know its easy, but I honestly cant figure it out. would appreciate if someone else can give this a look :)
😿
<@&286206848099549185>
<@&286206848099549185>
Its due in 20 mins please :(
every calcualtor i use says the same answer
what in the word am i doing wrong
strange trail
you found the right number, just have to round to the nearest hundred
you rounded to the nearest thousand
topaz oriole
you rounded to the nearest thousand
52,934,292.717
->
52,934,300
and its still wrong
i just used th wrong number
dude im so dumb
53050397
->
53050400
its correct
topaz oriole
yup!
i accdentally closed th tab and lost all my work. now the same old answer is FREAKING WRONG
what hte FLIP!!
53,050,400
is correct right?>
60,000,000 / 1.131 = 53050397.878
<@&286206848099549185>
@strange trail bro why is it wrong
WHAT THE FUCK
im so mad
strange trail
60,000,000 / 1.131 = 53050397.878
its not 1.131 anymore that number changed
topaz oriole
its not 1.131 anymore that number changed
wow
thanks
10 % = 1910
100% = 1910 * 10
19100 - 1910
= 17190
(WRONG)
can you maybe help me out again bruh @strange trail
strange trail
your calculation is correct up to subtracting 1910
he paid a tax of 1910, which is 10% of the original cost
so, the original cost is 19100, like you solved for
there's no reason to subtract off 1910 afterwards (in fact, to find the total cost including tax, you should instead add it to 19100)
topaz oriole
there's no reason to subtract off 1910 afterwards (in fact, to find the total co...
yeah thats correct, thanks yet again. you are turning into my hero dawg
strange trail
np good luck man
topaz oriole
I swear im better than these silly mistake, just super tired.
prime terrace
Can someone helppp?
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keen void
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keen void
omg i'm an idiot
split chasm
3/4 isn't 4/3 though
keen void
I take it back, i'm not an idiot
tough iron
I think its a mistake in the book though because i don't see a mistake anywhere its just addition
split chasm
did you copy down the question correctly / looking at the correct section for the solution
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keen void
It's c)
split chasm
^ you didn't copy down the question correctly
tough iron
Gotta be careful
keen void
I thought that was just a lazy way to make everything under the square root
split chasm
length of the radical bar indicates what's being rooted
tough iron
See f)
bronze eagle
May I get help
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bronze eagle
