this is like some delicious kind of chain rule
#help-39
honest spear
Or write dy/dx in terms of dy/dt
is what i was going for 
honest spear
say x = ln t we have $\frac{dx}{dt}=\frac{1}{t}$
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
jolly parrotBOT
Nyxzore
rough forge
i saw that
autumn thunder
say x = ln t we have $\frac{dx}{dt}=\frac{1}{t}$
$\frac{dy}{dt} = \frac{dy}{dx}...
but we dont know dy/dx
jolly parrotBOT
Nyxzore
autumn thunder
oh I get it now
honest spear
autumn thunder
thank you so much
honest spear
!done
pearl pondBOT
If you are done with this channel, please mark your problem as solved by typing .close
autumn thunder
say x = ln t we have $\frac{dx}{dt}=\frac{1}{t}$
$\frac{dy}{dt} = \frac{dy}{dx}...
one more question how do we deal with y
no derivative no chain rule
honest spear
y just stays the same
autumn thunder
y is in t we want to make it in x as in y(x)
isnt that what the question wants?
honest spear
holothink
abuse of notation at its finest just writing y fr
well i mean lets say you have y(t)
now you just have y(e^x)
then rewrite this as Y(x)=y(e^x)
but then you have to ignore my chainrule nonsense
autumn thunder
ok
rough forge
<@587715738484211754> might need help <:bearlain:762653539092529182>
What?
jolly parrotBOT
rough forge
Now use that do derive y''(t) as well
honest spear
the issue is we have ...-y(t)=0
rough forge
wdym
You have to express the equation in terms of dy/dx and d²y/dx²
y(t) becomes y(e^x) which by abuse of notation is just y(x)
technically speaking y(e^x)=f(x) some other function
honest spear
abuse of notation at its finest just writing y fr
.
pearl pondBOT
Available help channel!
stoic imp
say for example if I have a smooth curve, then any paranetrization of that curve is regular?
stoic imp
or is it like the opposite, if I have a regular parametrization of a curve then that curve is smooth
rough forge
say for example if I have a smooth curve, then any paranetrization of that curve...
You can have r(t)=(t²,t²) which represents y=x but r'(0)=(0,0)
stoic imp
You can have **r**(t)=(t²,t²) which represents y=x but **r**'(0)=(0,0)
so its not regular
stoic imp
You can have **r**(t)=(t²,t²) which represents y=x but **r**'(0)=(0,0)
what about r(t) = (t, t)
honest spear
say is a regular paramitisation $P(t)=(x(t),y(t))$ as star showed if the deritive is 0,0 somewhere it is not regular so we know the derivitive is non zero everywhere. so if we take some t=c and $x(c)=0 => y(c)\neq 0$
jolly parrotBOT
Nyxzore
honest spear
what about r(t) = (t, t)
i mean thats fine
but star showed a disprove by counterexample on your first statement
stoic imp
oh right, r'(t) = (1,1)
honest spear
say is a regular paramitisation $P(t)=(x(t),y(t))$ as star showed if the deritiv...
so clue where to go from here
stoic imp
.close
pearl pondBOT
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quasi depot
Hello, I was given this question to show that the limit exists. Although I'm having trouble with bounding the denominator. To be very honest our professor always uses the reverse triangle inequality to bound that, but I've always been confused by the initial assumption we make. Any tips?
dense jasper
Hello, I was given this question to show that the limit exists. Although I'm hav...
Suppose $\delta \leq 0.1$ or something and use that to bound $x$, then the denominator
jolly parrotBOT
Civil Service Pigeon
quasi depot
Suppose $\delta \leq 0.1$ or something and use that to bound $x$, then the denom...
Would 1 work or 0.1 is safer?
dense jasper
Try it
dense jasper
$-0.5$ satisfies $|x+1|<1$ but not $-2<|x|<0$. Why is there a $|x|$ there anyway?
jolly parrotBOT
Civil Service Pigeon
quasi depot
$-0.5$ satisfies $|x+1|<1$ but not $-2<|x|<0$. Why is there a $|x|$ there anyway...
Problem I had with using 0.5 was I had this
dense jasper
The lower bound of $-\delta$ is redundant because absolute values are non-negative
jolly parrotBOT
Civil Service Pigeon
dense jasper
I assume that’s messing you up
quasi depot
I assume that’s messing you up
Yeah.....been my issue since the start of the course really
Okay then from what we have, that doesn't look right....
pearl pondBOT
@quasi depot Has your question been resolved?
quasi depot
<@&286206848099549185>
halcyon estuary
Problem I had with using 0.5 was I had this
cant read the whole chat ... do u just need help for this range inequality?
quasi depot
cant read the whole chat ... do u just need help for this range inequality?
Yes, I'm trying to bound the denominator
halcyon estuary
first of all mod of x + 1 cant lie between -1.5 and zero
quasi depot
I didn't write that though
halcyon estuary
Problem I had with using 0.5 was I had this
range of x should between-1.5 and -0.5 for this
pearl pondBOT
@quasi depot Has your question been resolved?
quasi depot
range of x should between-1.5 and -0.5 for this
Really?
cosmic sinew
do you need help
broken moth
@quasi depot ok to show this you don't have to use numerical approximations
You just need to show the existence of delta
You can notice that X^2-2=(X-√2)(X+√2)
And |x+1|≤|x-1|+2
Or bruh
broken moth
Hello, I was given this question to show that the limit exists. Although I'm hav...
This is a false statement
It is not 1 it is 3
pearl pondBOT
@quasi depot Has your question been resolved?
high swallow
hi, could anyone give me a hint of where to begin? I'm kinda clueless on this problem. Thanks
smoky gull
are you familiar with modulo
true garnet
Use Chinese remainder theorem
pearl pondBOT
@high swallow Has your question been resolved?
high swallow
i'll have a think for a while i'll come back if i need more help, thanks
stoic imp
pearl pondBOT
stoic imp
[ 3. \text{ For each } n \in \mathbb{N}, \text{ find the possible values of } (3n^4 + 2n^3 - n + 1 : 3n^2 - 1). ] [ \text{ Provide an example for each value found.}]
jolly parrotBOT
Renato
pearl pondBOT
@stoic imp Has your question been resolved?
stoic imp
.solved
pearl pondBOT
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Available help channel!
cunning compass
did i do it correctly? 🙏
pearl pondBOT
did i do it correctly? 🙏
summer imp
Hummm you seem to have gone from 1-sin(theta) in the denominator to cos^2(theta)?
Amongst other things
still grail
for these establishing identity problems you should not be moving things from one side to the other
i think you tried doing that in step 2
cunning compass
oh
summer imp
Hummm you seem to have gone from 1-sin(theta) in the denominator to cos^2(theta)...
The identity is $1-\sin^\textcolor{red}{2}}(\theta) = \cos^2(\theta)$ for reference
jolly parrotBOT
Azyrashacorki
Compile Error! Click the 
reaction for more information.
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still grail
^^
sterile python
Yes you have mistaken it with Pythagoras identity
cunning compass
Hummm you seem to have gone from 1-sin(theta) in the denominator to cos^2(theta)...
oh mb forgot it has to be squared for pytha theoreom
i think i got lost somewhat
sterile python
Why are you multiplying 1-sinx/1-sinx
cunning compass
i thought getting the conjugate of one of the sides would give something
sterile python
1-sinx conjugate is?
cunning compass
oh
right
oh shit
i think i got it
so conjugate is mainly for using difference of square right
still grail
yup
pearl pondBOT
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fading nexus
How did I go wrong for 2
pearl pondBOT
How did I go wrong for 2
honest vigil
,rccw
jolly parrotBOT
honest vigil
is that 2/7 on the right?
fading nexus
Yea
honest vigil
why is it 2/7?
fading nexus
Crap was I supposed to subtract it
honest vigil
indeed.
fading nexus
Is this now correct? Also 4
honest vigil
,rccw
jolly parrotBOT
honest vigil
,calc ln(5/3)
jolly parrotBOT
The following error occured while calculating:
Error: Undefined function ln
honest vigil
,w ln(5/3)
jolly parrotBOT
honest vigil
I recommend prefacing decimals less than 1 with a 0 anyway. it was a little hard for me to see the decimal point through the half-erased writing.
and is the fourth digit there a 6 or an 8?
honest vigil
then it is correct.
I'll take a look at 4) now.
- is also correct minus the small nitpick I have about writing your multiplication on both sides under the equation/step (I would have preferred it written to the side).
but that's just a personal nitpick, so it's a pass.
fading nexus
Kk
Sorry for the rapid fire questions but for 1-18 am I just like switching the expressions
honest vigil
you can think of it that way if it helps you, yes.
fading nexus
Like this? And how would I find the decimal for 5?
honest vigil
,rccw
jolly parrotBOT
honest vigil
hint: rewrite 25 as a power of 5, and rewrite the cube root as an exponent.
for 2), yes.
fading nexus
hint: rewrite 25 as a power of 5, and rewrite the cube root as an exponent.
I’m not really sure how
honest vigil
25 = 5^?
fading nexus
How did you get rid of the root 3 tho
honest vigil
put it aside for the time being. we'll bring it back in in a bit.
fading nexus
It’s just 5 squared right so 2
honest vigil
correct.
so now we have a cube root over it.
do you recall what exponent a cube root corresponds to?
honest vigil
hint: a square root is the same as an exponent of 1/2.
fading nexus
So would it be 5/3?
honest vigil
1/3, not 5/3.
so a cube root is the same as an exponent of 1/3.
we can now use these two pieces of information to write the argument of the log using just exponents and a base of 5.
can you do it? (hint: recall that $(a^b)^c = a^{bc}$.)
jolly parrotBOT
Yukari
honest vigil
correct!
honest vigil
**Yukari**
now merge the two exponents using the rule here.
(which is called the power rule of exponents, in case you need a name to search it up later.)
fading nexus
2/5?
honest vigil
you mean $5^{\frac25}$?
jolly parrotBOT
Yukari
fading nexus
Yea
fading nexus
Wait 5^2•1/3
honest vigil
if you meant $5^{2 \cdot \frac13}$, that's right.
jolly parrotBOT
Yukari
honest vigil
so what will that end up becoming?
fading nexus
5^2/3?
honest vigil
absolutely correct.
so now you have $\log_5(5^{\frac23})$. can you continue from here?
jolly parrotBOT
Yukari
fading nexus
Wouldent I just stop there since I rewrote into the other expression and found x
honest vigil
oh yes, of course.
I was just going by your previous method and making it explicit if you want to.
but if you recognize this form and can stop here, then yes, this is good enough.
fading nexus
For this one how do I get rhe -4 to be postive since I can’t find a solution if it is negative?
honest vigil
is that $\ln(e^{-4})$?
jolly parrotBOT
Yukari
honest vigil
if it is, you already have the answer.
fading nexus
**Yukari**
I think
honest vigil
then you have the answer already.
fading nexus
No solution right
fading nexus
no solution?
honest vigil
there is a solution.
jolly parrotBOT
Yukari
honest vigil
the only things that cannot be negative are the base and the value of the argument.
but e^-4 is not negative.
fading nexus
Kk and sorry I got even more questions for 12 I’m not sure how to contuine and for 18 I don’t even know how to stat
honest vigil
for 12, recall how we handled the cube root earlier? the same strategy applies.
the same strategy for 12 kills 18 as well.
fading nexus
for 12, recall how we handled the cube root earlier? the same strategy applies.
...
Yea I tried doing that for 12 but I don’t know how to find 10^x=SR1000
I think mainly for 18 the form is confusing me
minor cloud
convert the square root into an exponent, and convert 1000 into a power of 10, like you did with the earlier problem.
oh, and don't forget to take into account that the sq. root is in the denominator this time round.
fading nexus
What would I do after this?
honest vigil
you're more or less correct, except you forgot to account for the fact that the square root is in the denom of the fraction.
fading nexus
I thought I got rid of the square root when I made it squared?
honest vigil
I mean getting a fraction of 3/2 as your exponent is correct. but the sign of that exponent is wrong.
fading nexus
Sorry I’m confused what you mean 😭
honest vigil
if I asked you to write 1/a using exponents, what would the exponent on a be?
equivalently, $\frac1{a} = a^?$
jolly parrotBOT
Yukari
fading nexus
A^2?
honest vigil
1/a = a^2?
hm, if we let a = 3, you're saying 1/3 = 9.
I'll spoil the surprise then, but you should review your exponents after this. $\frac1{a} = a^{-1}$.
jolly parrotBOT
Yukari
honest vigil
likewise, $\frac1{a^b} = a^{-b}$.
jolly parrotBOT
Yukari
honest vigil
so $\frac1{\sqrt{1000}} = \frac1{10^{\frac32}} = 10^?$
jolly parrotBOT
Yukari
fading nexus
How would I add that to the other expoinets just by adding a - ?
honest vigil
what do you mean?
fading nexus
Like would it be 10^-3/2
honest vigil
absolutely.
pearl pondBOT
@fading nexus Has your question been resolved?
pearl pondBOT
Available help channel!
fading nexus
For this one we have to find the domain of the graph and match it with its graph. And im having trouble figuring out the domain
fading nexus
Since I don’t know the original domain
livid aurora
Since I don’t know the original domain
Do you know where is log defined?
fading nexus
yea like I have these two graphs but I don’t know if that is still true for log10(x) like tiny 10
livid aurora
Yes
livid aurora
The more you increase the base of log the graph become less steeper
fading nexus
Ya
livid aurora
Is it the same domain for the natural log ?
It depends on the argument
Log base 10 (x+1)
Is defined in
x+1>0
x>-1
fading nexus
I don’t really know how to write domains like that we do like (0,♾️)
livid aurora
You can write x>-1 as (-1, infinity)
fading nexus
So all log equations have a base domain of (0,♾️)? When would we use the inverse version
livid aurora
So all log equations have a base domain of (0,♾️)? When would we use the inverse...
If log base a (x) is given
You have to make sure base a is not equal to 1
And x is positive
fading nexus
how would a base be equal to 1?
fading nexus
Sorry I’m just very confused because now I don’t know what to do with the + and - logs and ln
livid aurora
Sorry I’m just very confused because now I don’t know what to do with the + and ...
Do you know properties of log?
livid aurora
You can solve them by using these properties
torn flicker
You can solve them by using these properties
change base is just a bit more tedious compared to product and quotient rule
livid aurora
change base is just a bit more tedious compared to product and quotient rule
I just sent he/she all the property
torn flicker
I just sent he/she all the property
i know
livid aurora
Hmm
fading nexus
I haven’t done the properties yet so I don’t know how to solve with them
jolly parrotBOT
fading nexus
I just don’t know if the og domain changes when it is a natural log and I don’t know how to know if to use the inverse graph domain or the normal graph domain
Because I know log tiny2 (x) domain is the inverse but I don’t know what the other equations domains are
honest vigil
the domain of logarithms is always $(0, \infty)$ no matter what base you happen to be using.
jolly parrotBOT
Yukari
vivid grove
You can try experimenting with sketching logarithms on desmos. That may help in understanding
fading nexus
Im jusr paranoid cus we drew both these graphs in class and im worried i missed an equation for the other one
And I still don’t know how to find the domain because we don’t just have one log we are adding them together
Is the domain for LN also (0,♾️)
buoyant cypress
And I still don’t know how to find the domain because we don’t just have one log...
you can add the logs and convert into a single term then set the expression inside log to >0
fading nexus
I’m sorry I don’t know what you mean 😭
buoyant cypress
you have two terms log (x) + log (x+1) = f(x)
use the product rule here
fading nexus
We haven’t done product rule before
buoyant cypress
hm.. that is an issue then. you must learn the properties
fading nexus
I guess I’ll just leave it blank and ask my teacher in class it’s not worth the energy
Cus I have other problems to do
green aurora
you must intersect the domains
buoyant cypress
alr
fading nexus
For these one 14 I don’t know if I should multiply or divide and for 18 I’m not sure how to get the +2 out of the log
jolly parrotBOT
buoyant cypress
you must intersect the domains
that might not work always
vivid grove
For these one 14 I don’t know if I should multiply or divide and for 18 I’m not ...
It would be good if you provide context. I can't understand what you are trying to do?
buoyant cypress
it is kinda hard to read 😅
vivid grove
It would be better to type out your question and show working out
fading nexus
It would be good if you provide context. I can't understand what you are trying ...
Sorry the question is asking to solve the equation algebraically
vivid grove
What the equation
green aurora
that might not work always
it was refer to this case
fading nexus
What the equation
12,14,16,18 I mainly have questions on 16 and 18
vivid grove
Can you focus on one specific question
fading nexus
16 I guess it’s the more simpler question
I just don’t know if I should do loge(3)•0.045 or loge(3)/0.045
vivid grove
16 I guess it’s the more simpler question
So what is exact wording 16. I can't find the question
vivid grove
I just don’t know if I should do loge(3)•0.045 or loge(3)/0.045
Show working out from the very beginning
fading nexus
Sorry it’s a bad picture I don’t bring my textbook to school since I don’t have a locker
vivid grove
Much better
I would take ln not log.
It looks like you are using base e though so it does not matter
ln(e^(0.045x)) = 3
Edit: ln(e^(0.045x)) = ln(3)
By definition of ln we end up with:
0.0045x = ln(3)
From here it should be easy finish the proof the key understanding how logarithms are defined
fading nexus
I know I should put it in the calculator and like either divide or multiply but I do know which at least that’s how we did it in class
vivid grove
Forget about the original question. Suppose we have ax = b. How would we make x the subject?
fading nexus
Divide each side by a
vivid grove
We divide both sides by a. Because ax/a = x from equivalent fractions. Hence we get x = b/a
So what do we do in question 14?
fading nexus
Divide?
vivid grove
Correct
fading nexus
Ok and for 18 I don’t know how to get the +2 out from the log
harsh adder
Ok and for 18 I don’t know how to get the +2 out from the log
honestly theres no need to remove it
lemme see the q again
1s
flint basalt
,rccw
jolly parrotBOT
harsh adder
how bout
fading nexus
And there’s no like tiny number so I can’t do that
harsh adder
And there’s no like tiny number so I can’t do that
i didnt understand....?
fading nexus
I don’t know how to convert the equation because I don’t have a b value
honest vigil
if a base is not visible, there should be a line in your syllabus telling you what the default base is. usually it's 10 in algebra/precalc and e in calculus.
fading nexus
Oh ya
harsh adder
I don’t know how to convert the equation because I don’t have a b value
its not given so its 10
honest vigil
always double-check with your syllabus though!
fading nexus
This doesn’t seem right
harsh adder
keep these 2 in mind
- if no value is given its 10
- if ln is given then the base is e
fading nexus
harsh adder
This doesn’t seem right
no its wrong
fading nexus
I don’t know what I did wrong
harsh adder
because
honest vigil
there is quite more than one issue here.
fading nexus
From the other side ?
fading nexus
Huh
harsh adder
From the other side ?
and also btw if you have 10^-2 for instance you cant divide it by 2 and make it 5^-2
harsh adder
nope
fading nexus
It would be 12
harsh adder
nope
fading nexus
I’m so confused
fading nexus
8?
harsh adder
10^-2 is?
fading nexus
0.01
fading nexus
X=-1.99?
harsh adder
yep
fading nexus
Ty
harsh adder
Ty
you can only work on exponents with these rules
we used the negative exponent rule in that q
pearl pondBOT
@fading nexus Has your question been resolved?
green aurora
I don't understand when a force has a positive or negative sign
honest vigil
wouldn't it depend on what direction you've defined as positive?
green aurora
they are elastic forces
honest vigil
doesn't matter what kind of force or even vector they are.
green aurora
yes, but since the spring force is a restoring force, doesn't it always have a minus sign?
$F_e=-kx$
jolly parrotBOT
Mr. Penguin
green aurora
so if z is the direction positive
F1 =+ , F2=+ , and F3=-?
so kx+kx-kx
like this
bitter herald
I mean you can always define whatever coordinate frame you'd like
bitter herald
F1 =+ , F2=+ , and F3=-?
But this is ok
honest vigil
though with your choice of positive direction, wouldn't the signs be the other way around though?
perhaps I am misunderstanding something; if so I will apologize in advance.
bitter herald
The only important thing about the restoring force is that it is antiparallel (i.e., opposite) to the velocity vector. Whether it is mathematically positive or negative is not meaningful ultimately
green aurora
It seems that I just need to consider the absolute value of -kx and then put the signs according to the diagram
thanks!
.close
pearl pondBOT
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Available help channel!
remote flume
I don't get how we get 2L -1w
pearl pondBOT
I don't get how we get 2L -1w
remote flume
.close
pearl pondBOT
pearl pondBOT
Available help channel!
cunning compass
Are there no solutions for this? I always end with sec (started from left)
smoky gull
!show
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
hasty saddle
start from right, 1 + sin^2 is cos^2, so RHS is just cos
divide by sin, you get cot + tan = cot
tiny egret
start from right, 1 + sin^2 is cos^2, so RHS is just cos
would be true if it was $(1+\sin^2(\theta))\cdot \sec{\theta}$
jolly parrotBOT
lyric
hasty saddle
no nevermind im wrong completely
would be true if it was $(1-\sin^{2}(\theta)) \times \sec(\theta)$
jolly parrotBOT
Johannes Kepler
tiny egret
yeah minus mb
tiny egret
Are there no solutions for this? I always end with sec (started from left)
put it all in terms of cos and sin, becomes simple from there
*when you remember that the original equation must be defined
hasty saddle
just did that, you get $\cos(\theta) = 1$
jolly parrotBOT
Johannes Kepler
cunning compass
heres the thing i did
hasty saddle
$\sin(\theta) \cdot \left(\frac {\cos(\theta)} {\sin(\theta)} + \frac {\sin(\theta)} {\cos(\theta)} \right) = 1 + \frac {\sin^2(\theta)} {\cos(\theta)}$
jolly parrotBOT
Johannes Kepler
hasty saddle
then multiply in
hasty saddle
LHS becomes $\cos(\theta) + \frac {\sin^2(\theta)} {\cos(\theta)}$
jolly parrotBOT
Johannes Kepler
cunning compass
**Johannes Kepler**
wait if u dont mind me asking howd u get 1? wouldnt it be cos/1 as sin cancels
hasty saddle
**Johannes Kepler**
after this
$\frac {\sin^2(\theta)} {\cos(\theta)}$ are on both sides, so cancel
jolly parrotBOT
Johannes Kepler
hasty saddle
then you are left with $\cos(\theta) = 1$
jolly parrotBOT
Johannes Kepler
cunning compass
so is there no actual solutions
hasty saddle
no, $\theta = 0, 2\pi, \dots$
jolly parrotBOT
Johannes Kepler
hasty saddle
no you are correct
these are solutions, but cot is undefined at these angles
so no solutions
e.g. $\cot(0) = \frac {\cos(0)} {\sin(0)} = \frac {1}{0}$, undefined
jolly parrotBOT
Johannes Kepler
pearl pondBOT
Available help channel!
versed canyon
Can anyone help me with mcqs
pearl pondBOT
Can anyone help me with mcqs
pearl pondBOT
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.
pearl pondBOT
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versed canyon
And again, !nopdf
flint basalt
you can send screenshots of a problem you're stuck on.
however dont send downloadable files like explained above
versed canyon
I have answer keys for these but I want someone to explain & let me understand these
<@&286206848099549185>
summer imp
!15m
pearl pondBOT
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.
summer imp
And also, which ones in particular?
versed canyon
And also, which ones in particular?
All
versed canyon
No
versed canyon
Where do you get stuck?
In every mcqs
summer imp
Well you should try them first
And see where you get stuck
Otherwise we can't just know what you don't understand.
And with 100MCQs we're not exactly out of the woods...
Or at least the first few ones. The whole thing seems to cover a significant portion of a typical multivariable calculus course, so you should've already seen stuff pertaining to domains, ranges, partial derivatives, critical points, tangent planes, local extrema, constraint optimization, double integrals, change of variables, change of order of integration, Green's Theorem, triple integrals, Gauss' divergence theorem etc.
pearl pondBOT
@versed canyon Has your question been resolved?
errant badger
In every mcqs
digital summit
i have a feeling that AC is 10, but does anyone know why?
pearl pondBOT
i have a feeling that AC is 10, but does anyone know why?
digital summit
OH
dense jasper
pearl pondBOT
dense jasper
thank you my man
shut hull
<@&286206848099549185> sorry I am not patient and I'm getting sleepier by the minute
blazing garden
<@&286206848099549185> sorry I am not patient and I'm getting sleepier by the mi...
Bro….. do not murder me
Dexter…..
shut hull
Bro
My final
Is on
Thursday
This is the last topic
The final is worth 60% of mg grade
In so tired
I've been studying all day
blazing garden
I've been studying all day
So what do you need help with?
pearl pondBOT
@shut hull Has your question been resolved?
pearl pondBOT
blazing garden
@shut hull, I hope how I explained it will help
shut hull
.reopen
pearl pondBOT
✅ Original question: #help-39 message
shut hull
<@904377933144141894>, I hope how I explained it will help
Nah bro
blazing garden
Nah bro
I don't mean this as an insult, but do you know what a Taylor Polynomial is? Cause if not, that's smth we should go over
pearl pondBOT
@shut hull Has your question been resolved?
pulsar escarp
How do I get R? I know it's about solving for x with the other side being 1 but no combination I can think of gives a right fucking answer
pulsar escarp
zoomed in
Nevermind I got it. Used symbolab because I'm a fuck up that can't not rely on calculators for everything. I still don't understand this shit and never will.
.close
pearl pondBOT
pearl pondBOT
Available help channel!
midnight haven
I need help with De Morgan's law on complement of intersection of two sets
midnight haven
I can't understand the logic
honest vigil
so you need help understanding $\neg(A \land B) = \neg A \lor \neg B$?
midnight haven
x : x belong to (A intersect B)'
x : x does not belong to (A intersect B)
how do we come with the conclusion that
==> x : x does not belong to A or B
jolly parrotBOT
Yukari
honest vigil
sorry for the LaTeX mixup.
honest vigil
x : x belong to (A intersect B)'
x : x does not belong to (A intersect B)
how d...
if I restate that last line as "x does not belong to A, or x does not belong to B", would that make sense?
midnight haven
I can;t understand how it came to that conclusion
honest vigil
so let's think about it.
if an element x is in both A and B, then x is in their intersection.
agreed?
midnight haven
yes
midnight haven
no because x is strictly A
honest vigil
excellent, we're getting there.
midnight haven
ok
midnight haven
x is strictly in B
honest vigil
nicely spotted. and obviously if x is in neither A nor B, then x wouldn't be in the intersection.
midnight haven
that is symmetric difference of A and B
honest vigil
but we now see that to disqualify an element from being in the intersection, you only need to kick it out of one of the sets.
so if it is not in the intersection, then necessarily it is either not in A, or not in B. or both.
that's why if x is not in the intersection of A and B (equivalently, if x is in the complement of the intersection of A and B), then either x is not in A, or x is not in B, or both.
I hope it was clear.
midnight haven
why cant x be strictly A or B without the intersection
honest vigil
what do you mean?
midnight haven
why the shaded part not possible
honest vigil
not possible for what?
midnight haven
if x does not not belong to the intersection of A and B, the shaded part is possible too
honest vigil
does not, not?
midnight haven
ah sorry does not
honest vigil
did you intend the double negative or is it a typo?
right.
I never said it was not possible.
in fact, that's exactly what I mean when I said either x is not in A, or not in B, or both.
if you insist on expanding it fully, either x is not in A but in B, or not in B but in A, or not in both.
honest vigil
but De Morgan does not really care about what sets x is in with the way it is traditionally written, because if I say not in A or not in B, I'm leaving open the possibility of saying not in both.
and another reason for not being explicit (especially later on) is when you get to know the generalized De Morgan law (for any number of sets).
midnight haven
so he took it logically
honest vigil
you're not going to want to sit there and state (in A and B and C and ... BUT not in Z, or in A and B and C and ... BUT not in Y, etc.).
honest vigil
so he took it logically
yes, you can also view this from a logical perspective.
honest vigil
**Yukari**
which is this way of writing.
but ultimately, it's the same concept applied to different objects.
the idea is always that if an element is not in the intersection of some number of sets, then it must be OUT of at least one of those sets.
midnight haven
woah
alright I will go with the possible explanation that x is not in A or B, concluding x is totally outside of the circles
honest vigil
what do you mean, x is totally outside?
midnight haven
honest vigil
in fact, be careful with how you word this part:
x is not in A or B
notice that I worded it as x is not in A, or not in B.
midnight haven
x is not in A or x is not in B yeah
honest vigil
your wording implies $x \not \in (A \cup B)$. my wording implies $(x \in \overline{A}) \cup (x \in \overline{B})$.
jolly parrotBOT
Yukari
honest vigil
De Morgan concerns itself with the latter.
midnight haven
so lets start once again, I mean we came to the conclusion that x does not belong to A intersect B
therefore, x does not belong to A and B, am I correct
honest vigil
correct, provided you take that 'and' logically.
as in, x does not belong to A and B at the same time.
midnight haven
x does not belong to A and x does not belong to B?
honest vigil
now if you word it this way, that 'and' is an 'or'.
midnight haven
oh I see
honest vigil
but we now see that to disqualify an element from being in the intersection, you...
remember what I said earlier.
or, if it helps visually...
if x is not in the red area, then necessarily it must be outside the red area. but there are three places it can be: the green area, the blue area, or the yellow area.
midnight haven
Yes this
honest vigil
if you agree, then we can say that if x is not in the intersection (red area), then it is either out of A (placing x in either the blue area or the yellow area), or it is out of B (placing it either in the green area or the yellow area).
honest vigil
and that is the gist of De Morgan.
midnight haven
what
honest vigil
hm?
midnight haven
what point does he try to make then
honest vigil
think of it this way.
suppose you have a compound sentence, and you claim that not everything in that sentence is true.
more concretely, let's say someone says "oh A and B and C and D..." and your claim is that not everything in that sentence is true.
midnight haven
ok
honest vigil
now, "not everything is true" sounds like a hard claim, especially hard to prove. but because of De Morgan, you can turn this claim into an easier one: "at least one of those things is false".
isn't that much easier to attack?
instead of seeming like you need to show everything is false, you now only need to show one of them is false, and the whole sentence crumbles by De Morgan.
midnight haven
now, "not everything is true" sounds like a hard claim, especially hard to prove...
okay yeah
okay i see
hitting one except all
honest vigil
it's like a domino effect.
if someone says "A and B and C..." and you tip one of the dominos, the whole thing crumbles.
though it is worth noting that De Morgan's laws come in two parts, and this is but one of them.
midnight haven
so how is it related to this
honest vigil
well suppose then you want to prove that an element is not in the intersection.
midnight haven
yes
honest vigil
by its nature this would be rather hard to prove, but if you invoke De Morgan, you can now target any one set involved in the intersection and prove that that element is not in just ONE of the sets.
instead of having to consider every single one of them.
midnight haven
please re elaborate this
you can now target any one set involved in the intersection and prove that that element is not in just ONE of the sets.
instead of having to consider every single one of them.
honest vigil
you want to show that an element is not in all of some number of sets at the same time. at first, you might think we have to somehow show that it is not in every set one by one.
De Morgan says that you can do the same thing by just showing that it is not in just one of them, because then it wouldn't be in all of them at the same time.
midnight haven
what does 'in just one of them' mean
honest vigil
suppose you want to prove that an element is not in A, B, C, and D at the same time (that is, their intersection).
then, it suffices to prove that the element is either not in A, not in B, not in C, or not in D. if the element is not in as much as a single one of them, then surely you agree it can't be in the intersection.
midnight haven
wait a neither nor condition?
honest vigil
or maybe more concisely, if you want to prove that an element is not in the intersection of A and B, then it suffices to prove that it is either not in A, or not in B (basically, it is not in one of the two sets).
this can be generalized to any number of sets, but you're more focused on two sets, so let's stay there for now.
midnight haven
exactly, because multiple sets make me think harder
honest vigil
it's technically the same idea, and maybe you'll see it later.
honest vigil
or maybe more concisely, if you want to prove that an element is not in the inte...
but if you want the two-set version, this is it.
midnight haven
and another condition maybe that the element is not in both
so ultimately whats the conclusion
honest vigil
that is handled by stating that it is either not in A, or not in B.
the conclusion is De Morgan. there's nothing more to it.
at least from my point of view. unless you want to generalize this, which is up to you.
midnight haven
that is handled by stating that it is either not in A, or not in B.
is this phrase "either not in A or not in B" converted to set operations directly resulting in the conclusion?
honest vigil
yes.
honest vigil
either not in A or not in B, in set language, is x is in the complement of A, or x is in the complement of B.
which is how De Morgan for sets is usually stated.
midnight haven
actually my problem was
I was assuming more variables and possibilities whereares He did not
honest vigil
perhaps, yeah.
midnight haven
alright thank you so much!!
honest vigil
no worries.
honest vigil
'twas a good exercise.
midnight haven
tremendingly stressful
because I was just promoted to 11
with a class 10 brain, very difficult
honest vigil
we've all been there. you got this.
midnight haven
thank you..
I just realised a great mistake
A' means U - A = B right?
and vice versa?
for B'
@honest vigil
honest vigil
A' means U - A, correct, but U - A is not necessarily B.
honest vigil
U - B is any element of U not in B. this does NOT necessarily mean A.
some elements are part of neither set.
midnight haven
thats why I said U elements
framing more better
U - A = elements of U which do not contain A but contain B
honest vigil
also, "not containing A" is not exactly right. "not contained in A" is better.
midnight haven
I mean elements in U along with B without A
honest vigil
U - A = elements of U which do not contain A but contain B
U - A = elements of U that are not contained in A.
this statement does not imply membership of B.
midnight haven
wish my english was better
honest vigil
U - A should only talk about U and A.
midnight haven
and same idea for U - B
honest vigil
correct.
midnight haven
if I had taken these into account, I wuld have understood what De Morgan meant by saying "either not in A or not in B"
now I finally understood
Finally, thank you so much
honest vigil
no worries, glad to help.
midnight haven
alright I will close since I ned to bath ah
thank you for saving my hours
.close
pearl pondBOT
midnight haven
Prove that (A - B) U B = A
Given:- B (subset) A
My steps:-
x belongs to B => x belongs to A
(A - B) U B = {x : (x belongs to A and x does not belong to B) or x belongs to B}
Is this approach okay ?
bitter herald
Prove that (A - B) U B = A
Given:- B (subset) A
My steps:-
x belongs to B => x ...
Seems pretty solid to me
You should continue though
midnight haven
next step:
(A - B) U B = {x : (x belongs to A and x does not belong to A) or x belongs to B}
is everything okay here by logic?
bitter herald
next step:
(A - B) U B = {x : (x belongs to A and x does not belong to A) or x ...
Did you typo
bitter herald
You should backtrack a bit
bitter herald
Your conclusion is wrong to begin with
midnight haven
ok lemme see
bitter herald
Dors this make sense to you
What im saying is that this statement is a contradiction
It is never correct
midnight haven
I know therefore it becomes an empty set
but A - B is not gauranteed to be empty
so lemme re check
bitter herald
But it should be x does not belong to B you wrote A
midnight haven
I thought if x does not belong to B, should not with A as wel
but it isn't necessary
midnight haven
which
bitter herald
I thought if x does not belong to B, should not with A as wel
.
midnight haven
yeah
I think it should be
(A - B) U B = {x : (x belongs to A and x does not belong to B) or x belongs to A}
bitter herald
I think it should be
(A - B) U B = {x : (x belongs to A and x does not belong to...
Yes
midnight haven
So if union of a subset (A - B) with A
it results in A
because (A - B) is a subset of A
oh wait I got the answer
.close
pearl pondBOT
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old perch
Hi! I have a question.
pearl pondBOT
Hi! I have a question.
upbeat glade
Mhmmm.
old perch
I don’t know if I describe it clearly
This problem (Figure 1) reminds me of the Poncelet Porism (Discovered by a teacher himself). I’m not sure how to describe it clearly in words, so I drew a diagram instead. In Figure 2, α is a fixed angle, P is a fixed point, and Q is the moving point whose trajectory is a circle. Since the ratio PQ:PM is constant, we can obtain:
- The trajectory of the passive point is also a circle.
- Let O₁ be the center of the circle on which Q moves, and O₂ the center of the circle on which N moves. If the radius of circle O₁ is r₁ and that of circle O₂ is r₂, then we have r₁ : r₂ = PQ : PN.
- PO₁ : PO₂ = PQ : PN.
I already know how to prove the three statements above. Now I want to use the Poncelet Porism to solve the original problem (Figure 1).
In this problem, based on my drawing, I unexpectedly found that the locus of point F is an arc whose center is the midpoint of BC and whose radius is half of side A. The arc starts on side AB and ends on side AC. This seems to match property (1) of the Poncelet Porism.
The key question is: in Figure 2 there is one active point and one passive point, but in this problem there are two active points M and N and one passive point F. Does this situation still satisfy the Poncelet Porism? If it does, how can we use the porism to prove it?
First, r_O equals one half of r_B, which equals one half of AB, which equals a/2. (I feel that this follows from property (2), but at this moment I still cannot justify why r_O : r_B = 1/2.) Then, from property (3), we can obtain BO = 1/2 BC. (But again, just like the previous step, I still cannot justify why this is valid.)
If these two steps can be established, then we can prove that the locus of F is exactly that arc.
Is anyone watching? I really need help, thank you 🙏
bitter herald
Is anyone watching? I really need help, thank you 🙏
Your problem is highly detailed so help might not be available for a while. Either wait or see a relevant channel to post your question in in the 'Advanced Mathematics' or the 'Early University' categories
old perch
Your problem is highly detailed so help might not be available for a while. Eith...
Ok, thanks!
bitter herald
<@&268886789983436800>
pearl pondBOT
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old perch
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pearl pondBOT
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royal galleon
how does his last limit prove 6.27. The way he convers 6.27 it should be $\lim_{L(k+1)\to0} \frac{k}{L(k+1)}$ I don't see how proving the limit as k goes to 0 is related or how it helps
jolly parrotBOT
BigBen
pearl pondBOT
@royal galleon Has your question been resolved?
royal galleon
.solved
pearl pondBOT
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feral mantle
You're a teacher and a company offered you two profit models. Both profit models can get up from 4$ to 6$.
-First Profit Model: includes $4 per completed lesson and $2 per client no-show. If your performance rate (student retention) stays at 70% or more at minimum 7 clients for 3 months and tutor attendance rate is 85% minimum, your profit is increased to $5 per completed lesson and $3 per client no-show. If your performance rate (student retention) stays at 80% or more at minimum 7 clients for 3 months and tutor attendance rate is 85% minimum, your profit is increased to $6 per completed lesson and $3 per client no-show
- Second Profit Model:
Lesson Milestone Base Rate Retention Bonus Total per Lesson
1 – 10 $4.00 $0.00 $4.00
11 – 25 $4.00 +$0.50 $4.50
26 – 40 $4.00 +$1.00 \ $5.00
41 – 55 $4.00 +$1.50 $5.50
56+ $4.00 +$2.00 $6.00
For client no-shows, it stays only at $2.
Which model is mathematically more profitable? First or Second?
jolly parrotBOT
Draugr
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
feral mantle
You are a teacher and a company offers you two profit models. Both models allow earnings to increase from $4 up to $6 per lesson.
\subsection*{First Profit Model}
The first model is performance-based and defined as follows:
\begin{itemize}
\item Base rate: $4 per completed lesson and $2 per client no-show.
\item If the performance rate (student retention) is at least $70\%$ with a minimum of 7 clients for 3 months, and the tutor attendance rate is at least $85\%$, then:
\[
\text{Rate} = \$5 \text{ per lesson}, \quad \$3 \text{ per no-show}
\]
\item If the performance rate is at least $80\%$ under the same conditions, then:
\[
\text{Rate} = \$6 \text{ per lesson}, \quad \$3 \text{ per no-show}
\]
\end{itemize}
\subsection*{Second Profit Model}
The second model is milestone-based:
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\textbf{Lesson Milestone} & \textbf{Base Rate} & \textbf{Retention Bonus} & \textbf{Total per Lesson} \
\hline
1--10 & $4.00 & $0.00 & $4.00 \
11--25 & $4.00 & +$0.50 & $4.50 \
26--40 & $4.00 & +$1.00 & $5.00 \
41--55 & $4.00 & +$1.50 & $5.50 \
56+ & $4.00 & +$2.00 & $6.00 \
\hline
\end{tabular}
\end{center}
For client no-shows, the compensation remains fixed at $2.
Which profit model is more profitable?
jolly parrotBOT
Draugr
viscid shale
Important question, how much does a lesson last / take to do in time
feral mantle
Important question, how much does a lesson last / take to do in time
45 minutes per lesson.
15 minutes for the prepration.
Total 1 hour.
viscid shale
Any particular chance for no-shows?
feral mantle
No, but it will depend on the student if they decide to do no-shows.
viscid shale
Total 1 hour.
Id assume that we will consider a hard limit of 6 / 8 hours per day taken.
feral mantle
Yes.
viscid shale
Well, from a theoretical standpoint, 56 can be reached in about 10 days. Even if client amount increasing over time, id say you reach the peak of the second profit model in far less time
feral mantle
Id assume that we will consider a hard limit of 6 / 8 hours per day taken.
For every 25 lessons, there are 3-4 no shows.
viscid shale
Since for the First Model you have a 6 month cap that cannot be avoided.
Over an extended period of time, the first model is technically better in an optimal enviroment.
feral mantle
Well, from a theoretical standpoint, 56 can be reached in about 10 days. Even if...
I forgot to add, each client only goes for 2-3 lessons per week.
viscid shale
But considering the possibility of failing the 3 month condition for amount of clients, or simply a relative decrease of client noshows in the 2nd model
feral mantle
That applies on both profit models.
viscid shale
any cap on total clients?
viscid shale
If you get a good headstart you can prob reach peak amounts in the 2nd model in like 2 months.
And even if you dont, theres still a gradual increase to compensate for the time it takes.
feral mantle
But if the teacher has a goal of an long term bigger profit, would it be the same answer?
Since second one only has a stable 2$ no-show.
viscid shale
if for every 25 lessons you have 4 no shows theres a 4$ difference total
Havent particularly crunched the numbers, but i think that it would probably take a few years for 1st model to get back on the speed of 2nd model headstart.
feral mantle
Havent particularly crunched the numbers, but i think that it would probably tak...
Thanks for the answer. I really appreaciate it. 
feral mantle
Just out of curiosity, youre applying for a job i pressume
I'm already in the job, they offered me to switch. I'm working with them for 3 months now (1 month probation).
viscid shale
I'm already in the job, they offered me to switch. I'm working with them for 3 m...
Id guess ask coworkers about it, check how many have the 2nd option, etc...
And evaluate what has been your progress on this 3 months
feral mantle
Id guess ask coworkers about it, check how many have the 2nd option, etc...
So far, I will get stage 2 in a month now and some of my students are newly assigned to me (a month ago). And on my 1st month evaluation (I already passed the requirements so I'm just waiting for the 2nd month's email and currently performing on my 3rd).
feral mantle
Id guess ask coworkers about it, check how many have the 2nd option, etc...
On the other hand, the 2nd profit model got offered to everyone. Meaning, it is new for everyone.
viscid shale
if you got any friends inside the job and theres no NDA (or you dont really care about it, lmao), just talk about it, see what other think about it
like, the difference in payment is basically none
flint basalt
!occupied
pearl pondBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
feral mantle
.solved
pearl pondBOT
feral mantle
@flint basalt
flint basalt
<@674393799396556823>
yes? whyd you ping me
feral mantle
You want to use the channel?
flint basalt
no?
pearl pondBOT
Available help channel!
topaz burrow
hi ineed help in integral if its possible
pearl pondBOT
hi ineed help in integral if its possible
summer imp
Any specific one?
topaz burrow
those two
so
let me start with a if it was 1 then its arctan if 0 impossible
but if i need to complete than i need to talk a variable changement
and i couldnt know what it is
or if there is other cases
flat crane
multiply by 0 sonny
topaz burrow
0??
summer imp
Try to divide the numerator and denominator by sqrt(a).
olive stone
think trig substitution
summer imp
They already have the integral for arcsin, so dividing numerator and denominator by sqrt(a) and substituting is the easiest way to get back to that form.
topaz burrow
They already have the integral for arcsin, so dividing numerator and denominator...
but sub x with what ?
summer imp
Try to divide the numerator and denominator by sqrt(a).
Do that first.
topaz burrow
ok i did it
summer imp
If you distribute everything in the denominator you should get something that almost looks like the derivative of arcsin.
topaz burrow
ohhhhhh get it but like no metter a or i need to do her intervall
summer imp
What?
topaz burrow
a if it was 0 the form is false and if it was 1 then its artan
summer imp
You said that a can't be 0, it also can't be negative, so you can do this regardless.
topaz burrow
so i think in this case we need to do R positive not 0?
summer imp
I'm not sure what you're asking. For the denominator to even be defined you need to have a > 0.
topaz burrow
ok this is what i need thank you sooooo much
pearl pondBOT
summer imp
You too
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Available help channel!
craggy jay
If G is an abelian 2-group with distinct invariants, how can we construct an automorphism with the largest order? The order will be a power of 2 since the hypothesis implies Aut(G) is also a 2-group
pearl pondBOT
If G is an abelian 2-group with distinct invariants, how can we construct an aut...
@craggy jay Has your question been resolved?
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warped violet
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midnight haven
LINEAR ALGEBRA HELP NEEDED
pearl pondBOT
LINEAR ALGEBRA HELP NEEDED
midnight haven
23a and b make sense, I got A^2=I and confirmed M^2=I. 24 is where I am confused.
I will send my work one sec
24 still beign worked through but im confused
flint basalt
dense jasper
In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.
That being said, the dimensions here are inconsistent - remember that matrix multiplication is commutative. Idk if that is what's tripping you up.
But notice that the idea in 23a is ||a matrix with 1 and -1 on the diagonal can square to the identity||. If you need another pointer, consider ||blocks||.
midnight haven
thanks for letting me know
dense jasper
thanks for letting me know
pearl pondBOT
If you are done with this channel, please mark your problem as solved by typing .close
dense jasper
it's not closed lol
dense jasper
Closed messages look like what the Bot sent here ^
fluid axle
A = -D doesn't make sense, they ain't the same size
midnight haven
fuck....
fluid axle
CSP's first spoiler is a good idea here though
and it's essentially what you're trying to do with that "A=-D" thing
midnight haven
lord help me
idk what im doing
this is all I have
and even then im not confident
fluid axle
these would be the correct sizes inside M yes
the other interesting thing about 23(a) is that it doesn't matter what you put in the 3 spot, you'll still have A^2 = I
midnight haven
hmm
fluid axle
so the other hint I got is that you shouldn't give a crap about the C, pick only A and D so that M^2 = I whatever C is
midnight haven
am I on the right track
now theres a reason why AC+CD=0
hence why i said A=-D
or like
now it's partitioned and not just assumed
fluid axle
also you haven't corrected the error CSP told you in the first place, should be CA + DC but w/e
midnight haven
why does that order matter
ohhhh
fuck
DONT ANSWER THAT
listen... im new to this...
so basically ur saying
because positions 1 2 and 4 are I, 0, and I, whatver is in position 3 is irrelevant
ohhh
wait
bruh
C cant be 0
fluid axle
nah, the idea is that with a very particular choice of A and D you can make C irrelevant in the computation of M^2
midnight haven
i dont get it
fluid axle
think simple stuff
midnight haven
fluid axle
diagonal A and D
midnight haven
this all i got
fluid axle
diagonal A and D
there ain't a ton of diagonal matrices with A^2 = I or D^2 = I
midnight haven
am i supposed to be recalling something
fluid axle
nothing special no
pearl pondBOT
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.
midnight haven
shut your egg ahh up
pearl pondBOT
@midnight haven Has your question been resolved?
midnight haven
Being rude to ppl is a great way to discourage others from helping you...
rude where?
i was talking to the robot
slow oak
rude where?
it came across as if it was aimed at whoever triggered the bot
midnight haven
it came across as if it was aimed at whoever triggered the bot
Don't assume, I don't like robots it's that simple.
thick scroll
it came across as if it was aimed at whoever triggered the bot
I think the bot might automatically trigger it maybe if it detects someone pinging twice in a row? because I've noticed the bot doesn't delete the message command if someone triggers it
That being said I do agree with the sentiment of what you said that what was said was mildly disrespectful towards that rule
slow oak
It was triggered here, it doesn't auto trigger, often a few of the regulars will send delete commands for the bot because they don't want to deal with arguing with the person over the server rules
thick scroll
Ahhhh
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
wooden flare
how do I know if a point is on a line parallel to an axis in 3d?
honest vigil
if you have the equation of the line, sub the coordinates of the points into the line and see if the equation is satisfied.
wooden flare
uh something like this:
P(2, 3, -1) is apparently parallel but idk how and idk what other points are parallel to the z axis
honest vigil
I think the question is asking that what point can you join to P with a line such that the new line is parallel to the z-axis?