#help-0
keen plinth
vapid shuttle
This is lovely guys but I feel like a little off track
keen plinth
we can certainly apply ratio test here
worn fox
i think i was under the impression the ratio test was telling austin this, i see
vapid shuttle
I think I was last at |x|<1
but I had yet to show that this converged
I had only shown that anything outside of that did not converge
keen plinth
continue from here
vapid shuttle
What is the n-th term test?
fallen verge
If lim n-> infty of a_n != 0, the series diverges
vapid shuttle
yea
tardy stag
vapid shuttle
logarithms don't exist yet
fallen verge
OH ROOT TEST
tardy stag
didn't know we were in the 1500s
fallen verge
we're all silly
fallen verge
No worky?
vapid shuttle
$|n! x^{(n-1)!}|$
ocean sealBOT
austinu
ocean sealBOT
austinu
fallen verge
nth root of n! approaches 1 right?
tardy stag
not sure about that
fallen verge
Wait how
ocean sealBOT
austinu
real gazelle
Oh gosh
vapid shuttle
and we continue until we can say $<x^{n^{k}}$
ocean sealBOT
austinu
fallen verge
x^(n^n) right
vapid shuttle
and since |x|<1 then this converges to 1/1-whatever
which is < 1
then it converges for all |x|<1?
vapid shuttle
x^(n^n) right
idk
this is just the idea that came to mind
I think it needs polishing
real gazelle
Comparison to the sum of n x^n might be easier
keen plinth
[ x^{n \times n!} < x^n ]
ocean sealBOT
fallen verge
Oh lmao
vapid shuttle
Okay
I think I have enough to piece together a coherent answer now
so I'll get to work on this myself now for a bit
thank you everyone
real gazelle
You got this!
fallen verge
Always a joy to jive at the party
lone heartBOT
@vapid shuttle Has your question been resolved?
vapid shuttle
Okay here is my work for both, if anyone could check them that would be great and I would appreciate it a ton!
vapid shuttle
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lone heartBOT
lone heartBOT
Available help channel!
zinc ridge
tacit arch
https://docs.google.com/document/d/1kw4TTXMCeBUdhQkHSkHeubawywr4NazyT1rGDyvXhwA/...
Just screenshot and upload
alpine sable
not really that... Bt I just want to know more about p-adics or something like that
infinite digits before the decimal point
to the left...
lone heartBOT
@zinc ridge Has your question been resolved?
zinc ridge
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gritty verge
is it d for this?
lone heartBOT
is it d for this?
lone heartBOT
@gritty verge Has your question been resolved?
gritty verge
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ebon torrent
I want to understand this ^
lone heartBOT
I want to understand this ^
random relic
there are a couple of ways to solve this. one is to use elimination. you can add equations together sort of like you would numbers. see what happens when you add the left side of the two equations to each other and set that equal to the sum of the right sides
random relic
sorry i wasnt clear
i meant add (-2c + q) and (2c + q) together and set that equal to (-12) + (16).
adding the left sides and the right sides
try to see why those would still be equal
ebon torrent
like this?
young finch
what
you can set both equations so that the q is isolated
and then set them equal to eachother
i think that is conceptually easier
hidden surge
try using substitution
ebon torrent
so -2c = -12 - q?
hidden surge
then you divide both sides by -2 to get the value of c
young finch
but then you have a fraction
ebon torrent
like this?
young finch
avoid fractions theyre harder to work with
hidden surge
forget the last image, lets go back to square one
hidden surge
yes
austere silo
I'll tell you something I guess
This is how I used to "imagine" this things when I started it
ebon torrent
is there a way you can draw for me the next step ?
hidden surge
is there a way you can draw for me the next step ?
i'll gladly do it
ebon torrent
I understood algebra so far until I reached this question
austere silo
since in both equations there are q
You need to see other things
And difference one left side will be difference on the right side
Wait did y'all solve it
?
ebon torrent
no I haven't
hidden surge
austere silo
And you just get 4c=28
austere silo
-2c+q-(2c+q)=-12-18
hidden surge
what do I do after this
would recommend you to solve for q, because if you try to solve for c you would have to also divide the other side by -2
austere silo
-4c=-28
4c=28 c=7 and you calculate q than
that's probably easiest way because there are q in both equations
ebon torrent
I meant harder than what aioap wrote
austere silo
Sure
hidden surge
-2c + q = -12
2c + q = 16
that's your system of equations with two variables right?
ebon torrent
this is what I understood
hidden surge
yeah you're right
ebon torrent
the site says im wrong
hidden surge
because you're not fully finished
there you have the value of -2c = -12 - q
but you need to get the value of c
and you're going to do that by dividing both sides by the number that c is being multiplied by
ebon torrent
so I do the same with the first equation but now I look for the value of C?
hidden surge
cant
hidden surge
you gotta go to another help channel, sorry
fair pelican
My bad my bad
ebon torrent
can I look for C like this?
hidden surge
you look for C by dividing both sides by -2
ebon torrent
so C = 6 - 2q?
ebon torrent
oh I mean -2/q
hidden surge
but remember that you're dividing q by -2
austere silo
-q
ebon torrent
now I just solve the equation by replacing the c and q ?
hidden surge
wait a sec
austere silo
If you can't divide by -2 that easily you can multiply by -1 and than divide by 2
It's easier I guess
hidden surge
you then take the other equation, (2c + q = 16) and replace c with its value: c = 6 + -q/-2
ebon torrent
oh so those 2 equations the C and the Q are the same values?
I saw them as 2 different math equations with different values
austere silo
oh so those 2 equations the C and the Q are the same values?
That's what system is.
ebon torrent
how do I type -2 divide -q/2 in the caluculator ?
austere silo
you want to calculate -q/-2?
hidden surge
you're multiplying, not dividing
ebon torrent
I need to study for an exam in a different language and its a bit confusing
oh right im multiplying mb
hidden surge
yes
but you will need the value of q to solve for c
that's why we're doing this
just stick with me
here
you will remove -12 and -5 from the q side to get its value
hidden surge
where did you get 5 from...
can you give me like 5 minutes to give you a full explanation
ebon torrent
yea thanks
I think I got the 5 from here
this should be a lot more simple to do , the site doesn't give hints on the review section of each new math subject
it gives hints I can click only on the recorded lesson itself
but I watched it twice and I still can't solve this equation
austere silo
can I just tell you my solution please
ebon torrent
give me 2 min I think im solving it
austere silo
Okay good luck
ebon torrent
is this the actual way?
because 2c-2c is 0 and not C
its like 0= 28 thats probably not right at all
hidden surge
is this the actual way?
Nope
So your system of equations looks like this:
-2c + q = -12
2c + q = 16
You need to solve for c and q
Substitution is the easiest method
First you need to take one of the equations, it doesn't matter which one. In this case, I chose:
-2c + q = -12
You can either isolate c or q, but always try going for the ones that aren't being multiplied. In this case: q
Then, to isolate q you're going to remove -2c from its side of the equation using addition, since it's a negative number (subtraction when the value is positive)
-2c + 2c + q = - 12 + 2c
When we simplify we get
q = -12 + 2c
Now that we have the value of q, we apply it on the other equation
2c + q = 16
2c + (-12 + 2c) = 16
2c + 2c - 12 = 16
Now we simplify
4c - 12 = 16
Then we combine like terms
4c -12 + 12 = 16 + 12
Simplify
4c = 28
Then divide both sides by the number that our variable (c) is being multiplied by (4)
4c/4 = 28/4
c = 7
Now that we have the value of c we will use it to solve for q
Since q = -12 + c, that means:
q = -12 + 7
q = -5
austere silo
you could get q=-12+2c by dragging 2c from left to right side of the first equation
ebon torrent
I dont understand this part
shouldn't it be -24c + 4c = 16
or 2c squared
I think I understand now , I don't need to multiple by 2 because that is C and not Y
hidden surge
sorry, i made an error, let me explain again
To solve the given system of equations using substitution:
Equation 1: -2c + q = -12
Equation 2: 2c + q = 16
We can start by solving Equation 1 for q:
q = -12 + 2c
Now, substitute this expression for q in Equation 2:
2c + (-12 + 2c) = 16
Simplify the equation:
2c - 12 + 2c = 16
Combine like terms:
4c - 12 = 16
Add 12 to both sides:
4c = 28
Divide both sides by 4:
c = 7
Now that we have the value of c, substitute it back into Equation 1 to solve for q:
-2(7) + q = -12
Simplify the equation:
-14 + q = -12
Add 14 to both sides:
q = -12 + 14
q = 2
Therefore, the solution to the system of equations is c = 7 and q = 2.
ebon torrent
so I ignore the first solution you wrote and read this one?
hidden surge
yes, im so sorry, please excuse me
ebon torrent
ok I understand the first part finding c=7
hidden surge
Okay, but then?
ebon torrent
im doing the next step now
hidden surge
Okay so you know that
c = 7
and q = -12 + 2c
and now you're going to replace c with its real value
ebon torrent
yea this all makes sense thanks
I hope I can get another question like that so i can test myself
hidden surge
I hope I can get another question like that so i can test myself
i wish you the best, and sorry for making you go through this hassle
ebon torrent
I got this type of question following up so hopefully I can figure it out as well
thanks a lot
hidden surge
Good luck
lone heartBOT
Available help channel!
jade cliff
Can someone help me out with the c part?
lone heartBOT
Can someone help me out with the c part?
crisp cargo
what have you done so far
ocean sealBOT
jade cliff
why is lim x->0 cosx / x = 0
crisp cargo
$Res_{z=z_0} \frac{\phi(z)}{(z-z_0)^2} = \phi'(z_0)$
ocean sealBOT
chaac1
crisp cargo
why is lim x->0 cosx / x = 0
its not
jade cliff
oh i see
i was also a little confused about the concept of poles and singularities
so an isolated singularity is not a pole right?
crisp cargo
it is
crisp cargo
were you able to do parts a and b?
jade cliff
(cosx / x) has an indeterminate form at x = 0 right
yes i did them using this formula
crisp cargo
(cosx / x) has an indeterminate form at x = 0 right
well cos(0) = 1, so no
jade cliff
but 1/0 ?
oh
sorry
Okay so I apparently managed to do the first two parts using an incorrect method
But I got the answer so I didn't notice
crisp cargo
you should probably read up on residues
and make sure you understand this formula too
jade cliff
crisp cargo
i have to go, though
jade cliff
okay just one more thing
do u know any good resource for residues
my profs notes aren't really helpful as u can see
crisp cargo
one sec
honestly im not too sure
this is a good resource but it could potentially be confusing if you arent already thinking in terms of laurent series, here it is anyways though
tardy stag
it's kiiiinda cheating but I found this incredibly helpful in complex analysis:
http://davidbau.com/conformal/#cos(z)
it's also just really pretty
jade cliff
holy hell
tardy stag
is cosine
static kelp
is this sine and cosine?
jade cliff
thanks @crisp cargo and @tardy stag !!
jade cliff
how do i close the channel?
tardy stag
type .close
jade cliff
.close
lone heartBOT
static kelp
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steel olive
How to find the number of 6 digit numbers with 3 odd digits 3 even digits
steel olive
I did it by 5C3 ways to choose 3 odd digits
5C3 ways to choose 3 even digits
5³ ways to place 3 odd digits
5³ ways to place 3 even digits
Total = 100 × 5⁶
But answer is wrong
next elbow
I did it by 5C3 ways to choose 3 odd digits
5C3 ways to choose 3 even digits
5³ ...
You did the first part wrong
You should do 6c3 since it's a 6 digit number
And you should not repeat the same expression in the second step
steel olive
You should do 6c3 since it's a 6 digit number
But there are 5 odd numbers
next elbow
If you've picked 3 numbers of the 6 digits to be odd/even number, why would you still be picking 3 of 6 digits for the other set of numbers
next elbow
But there are 5 odd numbers
No, you should be picking the slots in the 6 digits to fill in the numbers
steel olive
No, you should be picking the slots in the 6 digits to fill in the numbers
So 6C3 × 5C3 × 5⁶?
next elbow
No wait sorry that was wrong
It should be 6p3
P as in permutation
Bcs if you rearrange the digits it would be a new number
next elbow
So 6C3 × 5C3 × 5⁶?
I apologize for not pointing out to use permutation but the second term is still wrong
jade cliff
Also you can't have 0 as the first digit, so you would have to subtract that case.
next elbow
Also you can't have 0 as the first digit, so you would have to subtract that cas...
Oh yeah but you can deduct it in the end
next elbow
So 6C3 × 5C3 × 5⁶?
It should be 6p3 * 3p3 * 5^6 - number of 5-digits
Bcs you're picking from the remaining 3 slots
Wait sorry i forgot those with repeated digits too
Maybe I'm bad at math
steel olive
Can we do 6C3 (positions for even digits)
That leaves 3 positions for odd digits
So 5³ ways for odd
6C3 × 5³ ways for even
So in totality 6C3 × 5⁶ × 9/10 (to account for the numbers starting with 0)
steel olive
Can we do 6C3 (positions for even digits)
That leaves 3 positions for odd digits...
Why cant 6P3 be used instead of 6C3 here
jade cliff
we are only choosing the digits here
next elbow
Why cant 6P3 be used instead of 6C3 here
My bad I was thinking wrong but since you can use the same number repeatedly there's no need to consider permutation
jade cliff
it doesnt matter if u choose (2,4,6) or (4,2,6)
because u will eventually arrange them
steel olive
Thanks a lot
lone heartBOT
@steel olive Has your question been resolved?
rocky glade
|x|-2 / x(x-1) > 0
will there be two cases here, x>0 and x<0?
vapid shuttle
$\frac{|x|-2}{x(x-1)}>0$?
ocean sealBOT
austinu
rocky glade
yea
vapid shuttle
yes, either x > 0 or x < 0
rocky glade
yes, either *x* > 0 or *x* < 0
so if x<0 then will it be -(x+2) in the numerator
vapid shuttle
yes
also, you should start thinking about now when the fraction will be positive. Either top and bottom positive, or both top and bottom negative. Those are the two scenarios we are looking for
rocky glade
oki thanks for yr help
rocky glade
also, you should start thinking about now when the fraction will be positive. Ei...
ahh, got it!
.close
lone heartBOT
lone heartBOT
Available help channel!
scarlet drum
We consider the triangle ABC, P a point located inside it and M,N,Q the means of the sides BC, AC and AB respectively. Let A', B, C' be symmetrical to the point P with respect to the points M, N and Q, respectively
a. Prove that quadrilaterals ABA'B', BCB'C', CAC'A' are parallelograms.
b. Show that lines AA', BB' and CC' are concurrent.
c. Prove that if at least two of the quadrilaterals in point a) are rectangles, then point P is the orthocenter of triangle ABC
rocky glade
We consider the triangle ABC, P a point located inside it and M,N,Q the means of...
atleast give a figure
lone heartBOT
@scarlet drum Has your question been resolved?
scarlet drum
<@&286206848099549185>
lone heartBOT
@scarlet drum Has your question been resolved?
west girder
i: use the fact that PBA’C, PAB’C, and PABC’ are parallelograms
ii: trivial by Jacobi’s theorem
I’ll try iii now
iii: use the fact that the reflection of the orthocenter of a midpoint is the antipode of the opposite side (APA’ is collinear, etc)
Refreshing to see geometry in this server
@scarlet drum
scarlet drum
i: use the fact that PBA’C, PAB’C, and PABC’ are parallelograms
ii: trivial by ...
we have to prove
that they're parallelograms
west girder
Yes… I’m giving hints
scarlet drum
hm
lone heartBOT
@scarlet drum Has your question been resolved?
cloud chasm
We consider the triangle ABC, P a point located inside it and M,N,Q the means of...
a)u can link M and N,MN is the median of a triangle of both triangle PAB and triangle PA'B',so AB is parallel and equals with A'B',so ABA'B' is parallelogram,for the same reason,BCB'C', CAC'A' are all parallelograms
cloud chasm
b)u can randomly choose a diagonal of these three parallelogram,the means point of this diagonal is the concurrent point of AA', BB' and CC'
u can easily prove it
c)because PO=OC',AO=BO,∠AOP=∠BOC',so triangle AOB and BOC' are congruent
then u can prove that AP is parallel with BC'
scarlet drum
c)because PO=OC',AO=BO,∠AOP=∠BOC',so triangle AOB and BOC' are congruent
Wait, where's O?
like what's O
scarlet drum
ah
cloud chasm
😂
then,u can assume that BCB'C' is a rectangle,that's nothing
so BC is vertical with BC'
and because AP is parallel with BC'
so AP is vertical with BC
cloud chasm
You mean perpendicular?
yes...
cloud chasm
It took me a while to translate that thing.
well at least u can understand
cloud chasm
so AP is vertical with BC
and for the same reason,whatever the next parallelogram u pick as a rectangle,BP will be perpendicular to AC or CP will be perpendicular to AB
so P is the intersection of two height line of triangle ABC
that means P is the orthocenter of triangle ABC
cloud chasm
@scarlet drum if u got it,u can close this channel
scarlet drum
.close
weary lake
lone heartBOT
weary lake
Need help withi this please
stuck loom
use quotient rule
weary lake
I did
I got root x x g'(x) - g(x) x 1/x root x
over sqr root x ^2
isnt the 1/2 from the the chain rule
1/2 (x)^-1/2
stuck loom
I got root x x g'(x) - g(x) x 1/x root x
there is a error in this
weary lake
whats the error
weary lake
static kelp
straight up quotient rule that shit
weary lake
yea thast what i did
whats the error
its not 1
🤣
wrong answer here
ignore that
stuck loom
whats the error
$f'(x)=\frac{\sqrt{x}.g'(x)-g(x).\frac{1}{2\sqrt{x}}}{x}$
ocean sealBOT
Nomad_InSearchfor_
stuck loom
$f'(x)=\frac{5}{2}$
ocean sealBOT
Nomad_InSearchfor_
trail wadi
lone heartBOT
naive crystal
First evaluate f(2)...
trail wadi
so does 2 substitude for x?
naive crystal
Yes...
subtle birch
what
trail wadi
7(2) -20?
subtle birch
no
naive crystal
2 >= 5??
subtle birch
x<5
trail wadi
so then 7(12)-20?
subtle birch
Yes
subtle birch
Yes
reef flame
How to find out the period of a periodic function from a graph?
tender anchor
do you have a specific graph in mind i can help you visualize how to find it
tender anchor
@reef flame I think your graph might be wrong
the function should go like a zig zag, from peak to through to peak to through
like going high to low to high to low
what you drew was a reflection of the left side
tender anchor
from x 7 -> 8 its right. but look from the point (-1,1) to (1,3) notice how it goes 2 to the right, then it hits the peak at (1,3) and goes down to (3,-1) moving 2 spaces to the right
when you look at your right sidde and you go from (4,1) to (5,3) you're only moving 1 space to the right
if u look closely its thinner than the left side you copied from
reef flame
Ok
tender anchor
@reef flame send a pic after u change it
reef flame
lone heartBOT
@reef flame Has your question been resolved?
coral vector
yeah i think that’s right @reef flame
tender anchor
ok so now to find the peiod its the distance from the start of the cycle to the end of the cylce, so the distance from one peak to the other
also the distance from one trough to the next
lone heartBOT
@reef flame Has your question been resolved?
lone anvil
This is a question asking poincare inequality in discrete case, for the subquestion d i dun get how to solve it, seems like the result from previuos subquestions are useful but i have totally no clue on this, its in numerical analysis topic
cursive lion
This is a question asking poincare inequality in discrete case, for the subquest...
Thats not a question
lone heartBOT
@reef flame Has your question been resolved?
midnight hare
Hey guys, may i ask how can i simplify this:
4 - Sqrt(2)*Cos(ArcSin(7/8)/2)
4 + Sqrt(2)*Sin(ArcSin(7/8)/2)
into this:
8 - Sqrt(15)
7
It really is like a magic to me lol
Thx!
tacit arch
let $\theta = \arcsin(7/8)$ then use half angle identities
ocean sealBOT
rie.mann
tacit arch
,tex .half angle
ocean sealBOT
rie.mann
tacit arch
and probably sin^2(x) + cos^2(x) = 1 somewhere
midnight hare
let me try it, thx you!
alpine sable
Hello how come x^2-19+78=0 turns into (x-6)(x-13)=0
lone heartBOT
Hello how come x^2-19+78=0 turns into (x-6)(x-13)=0
tacit arch
alpine sable
I'm confused how does this work
tacit arch
did you read this?
alpine sable
Yes
tacit arch
apply that to (x-6)(x-13)
alpine sable
i think i might have said it wrong the question
tacit arch
a = x, b = -6, c = x, d = -13
alpine sable
The book provides me with x^2-19+78=0
Is the final answer (x-6)(x-13)=0
I don't understand how they get to the final answer
wary stream
By factoring
tacit arch
yes i understand the question
tacit arch
apply that to (x-6)(x-13)
apply FOIL to this
alpine sable
okay one second let me try
tardy stag
to be clear, I expect your book actually provides you with
x^2 - 19x + 78 = 0
alpine sable
Yes
That is correct
i don't know how i mistyped that
i'm actually seeing what is happening now
(x-6)(x-13)=0
F x(x)=x^2
O x(-13) = -13x
I = -6(x) = -6x
L = -6(-13)= 78
all are coming back
-19x, x^2 and 78
=0
But how would I do this backwards, turn the first formule into this (x-6)(x-13)=0
As that's what the book wants me to do
tacit arch
By factoring
.
alpine sable
I don't understand how to
tardy stag
it requires some thought, i have a quick guide
alpine sable
i would appreciate that
tardy stag
i'm only going to deal with equations that are like x^2 + bx + c = 0, where b and c are integers
coefficients in front of x^2 make it more complicated
alpine sable
i have that sometimes 
tardy stag
okay, we'll get to that in a bit but it's simpler to start with this for now
alpine sable
Okay
tardy stag
actually let me write it as x^2 - Bx + C = 0 [note the negative B] scratch that, x^2 + Bx + C
the basic goal is to find two numbers that multiply to C and add to B
if i just put an example... x^2 - 3x + 2
because C is positive, we know those numbers are going to have the same sign
and because that middle number is negative we know those numbers are going to be negative
so you can write down (x - )(x - )
then you look for two factors of, well, 2, that add up to 3
this is to say, two numbers that multiply to 2
(there's only one possibility here)
tardy stag
yeah exactly so your expression would be (x-1)(x-2)
which, if you multiply that out as before [FOIL], will give you the original expression of x^2 - 3x + 2
alpine sable
so sometimes there's multiple possibilities to make it this format?
tardy stag
well, sort of... let's do another example
alpine sable
with different combinations of numbers
tardy stag
x^2 + 7x + 12 = 0
C is positive so we know our two numbers will have the same sign
and B is positive so we know they'll be positive
so (x + )(x + ) is our skeleton
now we write down factor pairs of 12:
1, 12
2, 6
3, 4
and we try each of these and ask, which one of these adds up to 7
alpine sable
3 and 4
tardy stag
right, so there were multiple options for the factors but only one of them worked for the sum
so (x + 3)(x + 4) would be the answer
alpine sable
Right
tardy stag
lastly, what if C is negative
alpine sable
that makes sense
tardy stag
x^2 - 5x - 24
since C is negative, our numbers will have different signs
so (x + )(x - )
and we make a list of factor pairs of 24, can you do that?
start with 1,24
tardy stag
that's all of them
vague lagoon
tardy stag
here we're looking for two numbers that have a difference of 5, since we know our signs are going to be different
alpine sable
so 8,3
tardy stag
yeah since our target is -5, we say -8, +3
vague lagoon
Good clip
tardy stag
please go away @vague lagoon
alpine sable
yeah since our target is -5, we say -8, +3
so it would turn into (x-8)(x+3)?
tardy stag
yes!
alpine sable
I am starting to understand it
but x would then either be 8 or -3 in that case then right
vague lagoon
<4+3i>
tardy stag
that's correct, x would be either 8 or -3 in that case
alpine sable
Awesome perfect this is great I'm understanding it
tardy stag
here you try one
x^2 + 9x + 18 = 0
tardy stag
yep good (and remember you can check by multiplying it back out)
alpine sable
Right
tardy stag
one more
x^2 - 10x + 25 = 0
alpine sable
I think (x-5)(x-5)
tardy stag
yeah! this one is slightly special because it's a square
doesn't really change anything but you could write it as (x-5)^2 if you wanted
alpine sable
i was a little confused with -10x + 25 but then i understand since -5*x=-5x so that would happen twice so -10x in total
but i was thinking double negative becomes positive but thats obviously not the case since it happens seperately
tardy stag
yeah the two negatives get added together so it stays negative
it does become positive for the third term though
okay do you want to try the more complicated ones, with a coefficient in front of x^2?
alpine sable
Yes
tardy stag
these are just more tedious
if we have Ax^2 + Bx + C = 0
we need to look at factor pairs of both A and C
and match them with each other, maybe flipping one around
to get to the target value of B
so starting simple again
2x^2 + 3x + 1 = 0
tardy stag
C is positive so our signs will be the same
B is positive so those signs will be positive
so our skeleton is ( x + )( x + )
yeah C only has the one factor pair of 1,1
so for this one you really have very few options, 1 goes after the plus sign (since it's the only choice) and 2 and 1 go before it
(2x + 1)(1x + 1)
let's try uh
2x^2 + 7x + 6 = 0
so same skeleton
factor pairs on the left are 1,2
factor pairs on the right are 1,6 and 2,3
and what we want to do is figure out which one of those factors (on the right) gets multiplied by 2 before they get added, to let us get to 7
it's a little clamplicated to put in words
but like our options are:
1 * 2 + 6
1 + 6 * 2
2 * 2 + 3
2 + 3 * 2
alpine sable
so 2+3+2=7
no wait th ats true but not the question
2*2=4+3
=7
This one it is 2 * 2 + 3
tardy stag
yep so let's look at our skeleton
( x + )( x + )
we put 2 and 3 on the right of the + signs, that's not hard to see i think
alpine sable
like earlier aswell
tardy stag
and since we're pairing the 2 with the 2, we put that in the opposite factor
so (2x + 3)(x + 2)
when you do the Outer term you get the 2*2
when you do the Inner term you get the 3
together they make 7
alpine sable
Okay that makes sense
tardy stag
that reversal is probably the hardest part of factoring trinomials like this, keeping track of where each factor pair goes
something slightly harder (and this is about as hard as you'd be expected to do i think):
6x^2 + 27x + 12 = 0
alpine sable
so for 6x^2 1,6 3,2 and for 12 3,4 2,6 1,12
tardy stag
mmhmm, now try and, like, pair up the pairs together to find something that adds up to 27
alpine sable
okay
tardy stag
so one possibility would be
1 * 1 + 6 * 12 which ofc won't work
alpine sable
hmm
i know what im supposed to do but testing them all is hard
but factors could also be turned around right
so 1,6 and 6,1
so that way you can change the order of the numbers to make it work
tardy stag
yeah it's kind of a pain
you only have to turn one side around at least (since if you turn both around you're back where you started)
alpine sable
Ooh
tardy stag
by the way I should say
this isn't the only way to do this, this is what I learned but there are other techniques including drawing a box or various diagrams
tardy stag
might help, I tried to stay under 12 but I wanted A and C to have multiple factors each
tardy stag
go on
alpine sable
Nope never mind this is hard
i had 3*3=9 but then it wont work out on the other side
13+64
1*3+6*4
no way
tardy stag
yeah it's a hard one
i'll narrow it down a bit, the target is odd so you know that at least one of the numbers is odd on each side
yeah that
tardy stag
that's right! can you write out the factored expression?
tardy stag
remember you can check your work with FOIL
tardy stag
easy to forget is that when we multiply numbers, they go in opposite factors
since what we're trying to do here is figure out the split between the Outer and the Inner term
so in this case it would be (x + 4)(6x + 3)
see how we have the 4*6x and the x*3
alpine sable
i had it turned around
alpine sable
see how we have the 4\*6x and the x\*3
Yes
tardy stag
there was actually an easier way to do that one
alpine sable
then it ends up with the good result
tardy stag
let's look at the original expression again
6x^2 + 27x + 12
what do you notice about all of those numbers?
tardy stag
particularly as it pertains to divisbility, factoring, etc
they're all multiples of 3
tardy stag
so what we can do is start by factoring 3 out of the entire thing, can you write what we get?
tardy stag
well, the 3 doesn't go away if we're just factoring
you might write 3(2x^2 + 9x + 4)
alpine sable
i thought we were allowed to do such action if we did it to everything
tardy stag
if we have an equation, yeah, but sometimes you'll be asked to just factor a polynomial on its own
if you do have a polynomial = 0 kind of situation then sure you can just divide everything by 3
alpine sable
or if its something that can be divided by 3 anyways right?:
so if on the otherside it would be 90
i could turn it into 30
alpine sable
Ah okay
I'm much better with the ABC formula
I can solve these much faster and more reliable is there any reason why we would want to do this
tardy stag
like the quadratic formula?
tardy stag
well, that's definitely an option and if you actually have an A that's not 0 it's sometimes faster
but once you do these enough you can get really fast at it
alpine sable
Do you know a place where to find exercises to specifically try these
tardy stag
and setting up a whole quadratic formula each time can get very tedious and slow
alpine sable
my book provided me with 5 of these which I all did
tardy stag
hmm, this is the sort of thing that I know there's lots of practice problems available, let me check my favorite site
here's a bunch of them, and they are separated into a = 1 and a not= 1
https://www.math-drills.com/algebra.php#solving-quadratic-equations-equal-zero
tardy stag
oh right that's a good question
you'll want a > 0 usually
and the easiest way to get that is to multiply everything by -1
(or, factor out -1 from everything, like we saw earlier)
alpine sable
you'll want a > 0 usually
and the easiest way to get that is to multiply everyth...
Yes i have been taught that
tardy stag
so if you started with -x^2 + 2x - 24 = 0
you'd rewrite it to x^2 - 2x + 24 = 0
alpine sable
Yes
tardy stag
what I would do is print out some of these and do them on paper
just feels nicer that way
alpine sable
Yes
tardy stag
oh yeah one last thing you'll probably run into
how do you think you would solve this?
x^2 - 49 = 0
tardy stag
yeah (that's the right answer but i'm going to talk about it anyway)
this has a special form called "difference of perfect squares"
and it factors like (x-7)(x+7)
if you run the FOIL on that you'll see that the Inner and the Outer cancel each other out
alpine sable
Hm so is it possible that they would force me on a test to use a specific method
i'm self studying and i pretty much got no idea what to expect and no teacher
I just got provided a book and the answers of the questions in the book and goodluck
to me it seems unlikely they would
tardy stag
in a classroom setting, you might be asked to use a certain method so that it's easier to see what you did wrong if you made a mistake
on an exam or whatever? no, that wouldn't come up
as I said earlier, the factor-pair thing is not the only way to solve these
it's just what I learned a, uh, while ago 
alpine sable
on an exam or whatever? no, that wouldn't come up
so I would probably be better off just to use ABC formula in case of doubt
tardy stag
nowadays i've done so many of them that most coefficients under 12 or so I can just do in my head
alpine sable
It's personally for me a safe bet
tardy stag
using the quadratic formula is a way to check your answers but again, it's pretty slow and honestly more error-prone since you ahve that square root
alpine sable
the answers it produces do show to me very fast if its correct or incorrect
tardy stag
and when you start talking about cubics, the ideas of factoring become more and more useful since the formulas are so painful
alpine sable
more often than not the answer makes no sense that comes out of it if it is wrong
i see
tardy stag
in fact, for equations with x^5 there is no formula for them
alpine sable
but you can't just take 5 times sqrt to make it go away
tardy stag
not just "we don't know one" there cannot be one
alpine sable
i didn't get that stuff yet
tardy stag
i mean like an equation like x^5 + 3x^4 - 2x^3 + 10x^2 - 3x + 1 = 0
anyway that's a long way out and there are special tricks for factoring those
here's something i used to do when i was bored
try factoring x^32 - 1
alpine sable
I have no idea how to
is it even possible
x^32=1
that's how far i will get
unless x =1
1^32=1
1=1
x=1
or -1?
i would say x=1 v x=-1
but it's probably not correct
tardy stag
remember what i said about factoring difference of perfect squares
and note that $x^{32} = (x^{16})^2$
ocean sealBOT
Hayley
tardy stag
and here I want you to factor it, not just solve the equation x^32 - 1 = 0
tardy stag
the general form for that DOPS factoring is like
$$a^2 - b^2 = (a-b)(a+b)$$
ocean sealBOT
Hayley
tardy stag
here let me start you out
$$x^{32} - 1 = (x^{16})^2 - 1^2$$
alpine sable
holy moly
ocean sealBOT
Hayley
tardy stag
forgot how to latex xD
alpine sable
i dont think i can honestly i have no idea where to go from this
tardy stag
so do you see we have (thing)^2 - (other thing)^2 here?
tardy stag
that means we can write it as (thing - other thing) (thing + other thing)
so here that would be $(x^{16} - 1)(x^{16} + 1)$
ocean sealBOT
Hayley
tardy stag
yeah it's one + and one - term
alpine sable
Right two options
alpine sable
(x^8-1)(x^8+1)
tardy stag
yeah so if we plug that into the original factoring, this one here #help-0 message what do we get?
tardy stag
no, we only showed that x^16 - 1 = (x^8 - 1)(x^8 + 1)
we didn't look at x^16 + 1 at all
alpine sable
no need i think
tardy stag
and yeah no need for extra parens there
alpine sable
no, we only showed that x^16 - 1 = (x^8 - 1)(x^8 + 1)
we didn't look at x^16 + 1...
Oh
tardy stag
yep
you could theoretically factor it if you started using imaginary numbers (these come from the square root of -1) but we don't usually care about solutions involving those
tardy stag
it takes practice ^_^
alpine sable
I'm catching up 1.5 years from the class
tardy stag
and curiosity
definitely do those practice problems I sent you, maybe like 1-2 sheets per day; don't go nuts on them
as you get more confident you can increase the difficulty
alpine sable
This only 1 chapter of the 2 chapters I will be tested on so 1-2 sheets per day is going to be very hard i think max 1 since i need to do another chapter
The second chapter I haven't even look at yet since I got 2 weeks left
tardy stag
well... i would argue that doing more is better
you want to get fast so you don't have to think about it
but i see what you mean
alpine sable
But it's crazy I got every 3 weeks a test to catch everything up
2 chapters at a time
normally they'd take 9 weeks
but that's the disadventage of switching to higher math
I'm going from normal math to higher math since i will need that for university but the school does not provide any teaching, only material and tests that i must pass
but i know what im doing it for
tardy stag
you can do it 💪
alpine sable
I hope so and believe so
THanks a lot for your help it means a lot and it helped a lot
I understand it now much better all it needs is more practice
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lone heartBOT
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Available help channel!
dim plaza
am i computing these correctly?
lone heartBOT
am i computing these correctly?
tender mango
Hi,
alpine sable
am i computing these correctly?
oh hi, berkeley numerical analysis coursemate
dim plaza
oh hi, berkeley numerical analysis coursemate
lmfao hello have u done this question do u wanna sanity check it
dim plaza
and I'm behind on lectures 💀
lmfaoo same i do the hw section corresponding to the lecture immediately after... so i havent watched 3.4 onwards
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rotund crater
what means a(1/a)=1
lone heartBOT
what means a(1/a)=1
alpine sable
well clearly a/a = 1 provided a isnt zero
long axle
A non-zero number multiplied by its reciprocal equals 1
lone heartBOT
@rotund crater Has your question been resolved?
dull lily
lone heartBOT
dull lily
Im doing activity 1!
so this is my first step
what im thinking right now is to multiply every single value in the u -> vector by 8, is that step correct
vale crag
hmm
harsh swallow
yes
vale crag
not sure where you're going here
maiden isle
what im thinking right now is to multiply every single value in the u -> vector ...
yes multiply by 8. then what happens?
harsh swallow
you're right in spotting that in order for it to be a linear combination of u and v the last component has to be 8 * the 1 in u
dull lily
is this step correct
maiden isle
why -5?
dull lily
if i multiply everything out
on 5v
i get the values 20, -15, 40, -25, 0
right
24 - 20 = 4
-16- -15 = -1
so on so fourth
i just need to know if im going in the right direction
maiden isle
what about the last component?
maiden isle
ask this question: is 8u + -5v that same as the desired vector?
tacit arch
is there more to this question
vale crag
it's a continuation of the previous thread
plain flame
alright so i want to bound |5 - z| from below
lone heartBOT
alright so i want to bound |5 - z| from below
plain flame
knowing that |z| <= 7/2
it seems reasonable to just write |5 - z| >= |5 - 7/2| = 3/2
but is that correct>
i think i should apply some triangle inequality but i have no clue which
vale crag
seems very reasonable
didn't you just apply a triangle inequality anyway ?
from wiki
@plain flame
tacit arch
you can be slightly more rigorous by letting z = a + ib and minimizing |5-z|^2
plain flame
you can be slightly more rigorous by letting z = a + ib and minimizing |5-z|^2
fair enough yes
ill do both
thank you
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topaz leaf
I have an alg question that gives me the parabola y = x^2
lone heartBOT
I have an alg question that gives me the parabola y = x^2
topaz leaf
and the problem is
Write an equation that shifts 2 units down and is reflected across the x axis
I know how to do each of them individually but I'm not sure if I do -y = x^2 and then do y = -x^2 and then do y = -x^2 + 2
or
y = x^2 - 2 first and then -y = x^2 - 2
keen mason
y = x^2 - 2 first and then -y = x^2 - 2
what would happend if you were to do it the other way round
topaz leaf
y = -x^2 + 2 or y = -x^2 - 2
sorry I typed the same case twice but I changed it my brain is fried today
cloud chasm
y = -x^2 + 2 or y = -x^2 - 2
will,it's decided by the order u take to this parabola
if 'shift' goes first,y = -x^2 + 2 is right
if 'reflect' goes first,y = -x^2 - 2 is right
lone heartBOT
@topaz leaf Has your question been resolved?
foggy current
The question is to find the length of the parametric curve $x=t-2\sin t,y=1-2\cos t$ where $0\leq t\leq 4\pi$. The integration part is no big deal but it has to be proven that the curve is only traversed once. I first tried differentiating $x$ and $y$ to try to show that their derivatives stay the same sign, but that didn't work.
ocean sealBOT
math_is_fun
young finch
you cant use a graph?
foggy current
I'm just curious if there's a non-graphical method
young finch
maybe show there are no points where the curves are the same distance from the graph at that particular radian
foggy current
What do you mean?
weary wyvern
I think you need to show r dot is never 0 where r parameterises the curve
foggy current
Is r the distance from the origin in polar coordinates?
weary wyvern
Is r the distance from the origin in polar coordinates?
No like r(t) = (x(t), y(t))
foggy current
Yeah I know
young finch
oh
twin nimbus
@foggy current if the curve is not traversed "once" that implies it's periodic. However, we can see from x(t) that this curve cannot be periodic.
This doesn't rule out self intersection, but it doesn't seem that you're concerned about that?
foggy current
This doesn't rule out self intersection, but it doesn't seem that you're concern...
Yeah that's fine
twin nimbus
(and of course there are pathological cases where a curve can retrace itself that are non-periodic, but this is not one of them)
weary wyvern
I think you need to show r dot is never 0 where r parameterises the curve
This is basically showing that x and y can't change sign at the same time, so it's close to what you were trying
foggy current
I think you need to show r dot is never 0 where r parameterises the curve
That doesn't hold
foggy current
<@1105989284592353330> if the curve is not traversed "once" that implies it's pe...
I think this will do
Thanks everyone!
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lone heartBOT
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Available help channel!
silent monolith
lone heartBOT
granite rapids
hi
silent monolith
ye
lone heartBOT
@silent monolith Has your question been resolved?
twin nimbus
Why is part of your image whited out?
tardy stag
that's where the answer would be
lone heartBOT
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Available help channel!
tender anchor
so when u have sqrt(x) it can be simplified to x^(1/2) w the formula rootb(sqrt(x^a)) = x^(a/b) which is just a simplifying method so i also have these 2 formulas written down:
nsqrt(x^m) = 1/(x^m/n) (n being the root not a constant)
is this formula the derivative of sqrt(x) or just simplifying sqrt(x)? is this formula even correct?
so when u take the derivative of sqrt(x) ik it is 1/(2sqrt(x)) whats the formula for that?
is it m/(n(sqrtx))? (m as the exponent of x in sqrt x, n as the value of the root)
tender anchor
im not sure when u use the formula nsqrt(x^m) = 1/(x^m/n) like what context thats used in
tawny condor
its not a derivative
its exponent rules
just a different way of writing the same thing
tender anchor
the 1/x^m/n?
tawny condor
yeah
tender anchor
so then writing 1/x^(m/n) is the same as writing sqrt(x) same as writing x^(1/2)
and the derivative rule of sqrt(x) is the 1/2sqrt(x) one where its m/nsqrt(x) right?
tender anchor
like when u do the derivative of sqrt x its 1/2sqrtx
tawny condor
this doesn't have anything to do with derivatives
wide marten
The derivative of a sqrt is just a power rule
tawny condor
like when u do the derivative of sqrt x its 1/2sqrtx
that is true yes, make sure to use parenthesis tho
tender anchor
im just getting my formulas mixed up
yeah
i have a derivative question also how is the derivative of 1/x = 1/(x^2) like how does that work
ocean sealBOT
rie.mann
tender anchor
o nvm it makes sense
tacit arch
use negative exponent with a=x and x = 1
shrewd ruin
I have a matrix question
lone heartBOT
I have a matrix question
echo socket
I don't knnow where to start
Do you know about the geometric meaning of the determinant?
lone heartBOT
@shrewd ruin Has your question been resolved?
lone heartBOT