could you solve these systems like youd solve any system of a set of equations
#help-36
dim flume
blissful meadow
Yes, although if by any you mean linear systems of equations in general they won't be linear systems.
This isn't a linear system for instance
But again a good strategy can be to solve for lambda in one of the more simple equations and plug that into one of the other.
Or you can work like this. Pick an equation, try to see how it could be satisfied, and investigate the cases.
dim flume
But again a good strategy can be to solve for lambda in one of the more simple e...
which method is easier 🤣
blissful meadow
The second one is less error-prone since you're not dividing by anything.
When you divide by any of the variables you have to check the case where this variable is 0
dim flume
Or you can work like this. Pick an equation, try to see how it could be satisfie...
okay so you mean this method right
blissful meadow
Yes
dim flume
whats the idea behind this method exactly
blissful meadow
For the system to be satisfied, you need all three equations to be satisfied, so you can pick one equation and check in which cases it would be satisfied.
Then any solution will satisfy one of those cases.
So for instance in this case, we see that if we're going to get a solution to this system, we'll need either r=0 or lambda*r = 0
If you start with the first case r=0, what do you notice with the third equation (the constraint)?
dim flume
If you start with the first case r=0, what do you notice with the third equation...
wait one more thing why did we introduce the constraint here and not previously
like in the plane problem i think we just dealth with the three equations that we got from the gradient right
blissful meadow
Yes I didn't bring it up then since the system was relatively easy
But you really had 4 equations and 4 variables (which is what you need to at least get a solution anyway)
dim flume
But you really had 4 equations and 4 variables (which is what you need to at lea...
so you introduce the constraint equation whenever you have more variables than you have equations?
blissful meadow
You always have n+1 equations in n+1 variables.
Like for instance for the plane you had, what you got from the three equations wasn't numbers. You got relations the inputs need to satisfy to be solutions to the system : x=z and x=-2y.
You needed the fourth equation to get actual numbers as a solution
dim flume
You always have n+1 equations in n+1 variables.
so when do you introduce the constraint and when do you not
blissful meadow
You always do
blissful meadow
Yes
dim flume
oh i see
blissful meadow
You already had 3
What you did in the end plugging in x=z=-2y was using the fourth one
You could've very well started with that fourth equation getting y = 2x + 2z - 16 and used that in the other equations
You would've gotten the same thing
dim flume
You already had 3
so i cant just use two equations here without the constraint
see whats confusing i guess is the method of solving for the system
blissful meadow
so i cant just use two equations here without the constraint
To solve a system of 3 variables you need 3 equations.
dim flume
To solve a system of 3 variables you need 3 equations.
ohh okay
so since we have 3 equations and 2 unknowns we introdoced the cpnstraint equation
blissful meadow
No, you have 3 equations and 3 unknowns
You always need to use the constraint equation
Just like you did in the end with the plane
At some point or another you'll use it
blissful meadow
Yes, but then really what you have is a system of 3 equations and 3 unknowns
dim flume
woah okay im confused, we have 2 equations and 2 unknowns, is that not enough
for the solution
its not even 3 unkowns
blissful meadow
lambda is an unknown.
We don't really care about its value, but it's still an unknown in there.
So 3 equations $f_x = \lambda g_x$, $f_y = \lambda g_y$, $g(x,y) = 0$ and 3 unknowns $x$, $y$ and $\lambda$.
soft zealotBOT
Azyrashacorki
dim flume
lambda is an unknown.
oh right 💀
so here we had 3 unknowns and 2 equations so we introduced the constraint right
if we had 3 equations and 3 unknowns we dont need the constraint to solve the system right
blissful meadow
You always have n+1 unknowns and n+1 equations
.
It's not a question whether or not we add the constraint.
You need the constraint to get a solution
Again going back to the plane earlier, x=z=-2y is not a solution. There's lots of points that satisfy this.
You need the constraint, always
dim flume
You **always** have n+1 unknowns and n+1 equations
oh alright i see now
alright so
im best at solving linear equations 🤣
what are good methods to solve non linear ones like in this case
blissful meadow
For the system to be satisfied, you need all three equations to be satisfied, so...
This
leaden moon
what are good methods to solve non linear ones like in this case
there are many kinds of system like that
blissful meadow
As you deal with more of them, you'll get the hang of what to do when
dim flume
Plug what you got in.
soft zealotBOT
Azyrashacorki
blissful meadow
This doesn't mean that $2-2\lambda r = 0$
soft zealotBOT
Azyrashacorki
blissful meadow
You need a product = 0 to conclude that
dim flume
oh shoot
blissful meadow
One trick is to try and look for equations that don't contain all the variables.
For instance, equation 2 has only 2 variables in it, so that means you may be able to solve for one of them in terms of the other.
dim flume
yeah idk what the heck im doing
blissful meadow
If you use equation 1 then you can still get something meaningful, but it's more complicated.
Use this
blissful meadow
Okay then you can use that in the other equation with lambda.
Also note that you would need to consider the case r=0 in that equation as well
dim flume
oh yeah
wait so
why dont we just solve the system using the substitution method
or is that what were doing here
yeah thats what were doing
i think im trippin
this is what too much math does to the brain 😂
i mean you know when solve for one variable
blissful meadow
It is yes. Again you could also just straight up solve for lambda in one equation and plug it in the other.
dim flume
plug into the other equation
blissful meadow
But in general this will involve dividing
And this means considering cases.
It's up to you.
dim flume
idk what im doing 🤣
blissful meadow
I mean you can do that but you already had lambda = 1/r
Why not just plug that in the first equation where you have lambda?
dim flume
Why not just plug that in the first equation where you have lambda?
good point
i should get it in terms of 1 variable
blissful meadow
good point
What relation do you get between r and h from there?
dim flume
What relation do you get between r and h from there?
i can solve for h and plug for the third equation?
or
blissful meadow
Well you have 2r =h
dim flume
in an equation where theres only h and r
blissful meadow
You can plug that inside equation 3 yes
dim flume
Well you have 2r =h
wait theres a + h
dim flume
What do you think I did to get 2r=h
oh probably some algebra
blissful meadow
Yes, you just subtract h from both sides.
Once you have that you can plug it into equation 3 and you’ll be able to solve for r
Must go now. You’re almost done
dim flume
Must go now. You’re almost done
thanks alot dude, really appreciate you taking your time, thanks alot
final saddleBOT
@dim flume Has your question been resolved?
tender pollen
!15m
final saddleBOT
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thorny tendon
hello, can i get help on this problem
!status
final saddleBOT
hello, can i get help on this problem
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
i dont know where to begin
i dont know what my constraint function is
nor my objectivve function
thorny tendon
it would start by labelling everything you can with variables: give quantities names so you can refer to them in the problem
dim flume
it would start by labelling everything you can with variables: give quantities n...
we know the volume of a box is 5*a^2 = 4ft^3
right
or actually
dim flume
6a*a^2 - 2a = 4ft^3
thorny tendon
i'm not sure what you've labeled a, but i think a picture to draw this out would be best
dim flume
i'm not sure what you've labeled *a*, but i think a picture to draw this out wou...
a is a side length
thorny tendon
side length of the base of the box, i assume
dim flume
well yeah
well any side
then all that volume subtracting the top from it
because theres no top
thorny tendon
i think you have to assume the base is a square, problem not worded well
dim flume
well its a box
thorny tendon
volume looks to be area of the base times the height
dim flume
so cube
dim flume
volume looks to be area of the base times the height
yes
thorny tendon
boxes can be rectangular, but in this case i think there are too many unknowns if you have base != height
dim flume
oh
thorny tendon
volume looks to be area of the base times the height
this is 4
and then you need to figure out what the total area of material you use to make the box
thorny tendon
volume = 4, so given your variables, what is your expression for the volume = 4?
base * height = 4, and using variables, what does this eqn look like?
dim flume
volume = 4, so given your variables, what is your expression for the volume = 4?
6a^2 = 4?
thorny tendon
what is 6 in this case?
dim flume
the sides
thorny tendon
oh lol
dim flume
a^3 = 4 😂
dim flume
idt the box height = the base side length
your saying it may be rectangular?
in that case
whl
W x H x L = 4
thorny tendon
what i'm saying is that an open box with a square base may not be exactly as tall (height) as it is wide (side length)
i.e. a != h
dim flume
what i'm saying is that an open box with a square base may not be exactly as tal...
yes
so that would be a rectangular prism
what am i missing
thorny tendon
what i am saying is that
V = A * height = a^2 h
so V depends on side length and box height
dim flume
what i am saying is that
V = A * height = a^2 h
alright
area times whatever hight may it be
sure
so thats our objective function right
thorny tendon
i think so. i never learned objective function, but this is what we'll use to get a in terms of h or vice versa
We need the area of the material used to make the box
dim flume
what i mean by objective is the function being restricted
not the one restricting
whats the restricting one
<@&286206848099549185>
dim flume
could you please elaburate on this
damn bro just deleted
That’s crazy
Can anyone help me out on this
<@&286206848099549185>
candid pulsar
hello, can i get help on this problem
is it given that the cardboard box is made out of a cut square plane?
not all boxes can be made this way
then assume the side of the large square is a and squares of side x are cut from each corner
dim flume
is it given that the cardboard box is made out of a cut square plane?
idk i just wrote down whatever tbh
what should i do
candid pulsar
can you find the dimensions of the cuboid you get from this plane then?
meaning its length, breadth and height
lean granite
?
dim flume
can you find the dimensions of the cuboid you get from this plane then?
l^2 *h?
dim flume
yes but what is it in terms of a and x
why do we need to write it like that?
candid pulsar
were folding the sheet to get a cuboid
dim flume
i dont really get it
lean granite
What is the actual question
candid pulsar
hello, can i get help on this problem
...
dim flume
😵💫
candid pulsar
hmm you dont seem to be given enough context but the diagram you attached with the question implies youre making the box out of a square sheet
dim flume
i mean i just drew that for the sake of trying something
idk what youre even supposed to consider
proud fern
can you find the dimensions of the cuboid you get from this plane then?
@dim flume
did u understand this
dim flume
<@1402083806034268161>
Yes thanks, so is that our objective function or our constraint one
I believe it’s the objective
Right
So what’s constraining it
proud fern
So what’s constraining it
hold on a sec
proud fern
So what’s constraining it
i think this approach is not good
it might be not be able to solve the question
my previous approach was better
dim flume
I understand the formula now, but how do we proceed with the optimization part ?
proud fern
I understand the formula now, but how do we proceed with the optimization part ?
can u tell what topic is this question from
it seems it involves differentiation
dim flume
it seems it involves differentiation
Yes it does
dim flume
have u learnt about maxima and minima
Yes
dim flume
The surface area is the constraint function ?
proud fern
did u understand the solution
final saddleBOT
@dim flume Has your question been resolved?
final saddleBOT
halcyon ether
<@&268886789983436800> get themmmmmmmmmmmmmm
final saddleBOT
halcyon ether
worldly spruce
final saddleBOT
Available help channel!
sonic crystal
Hi for part b when it asks to write the double integral is it asking me to write in cartesian or polar
final saddleBOT
Hi for part b when it asks to write the double integral is it asking me to write...
sonic crystal
or is either ok
drowsy epoch
I'd say either, since not specified
final saddleBOT
final saddleBOT
Available help channel!
gentle zephyr
final saddleBOT
gentle zephyr
\textbf{Exercise 4.} Let $a, b \in \mathbb{Z}$, not both zero, such that $5 \mid a$ and $\gcd(a, b) = 7$. Calculate $\gcd(2a^2 - b^2, a^2 - 3b^2)$.
soft zealotBOT
Renato
leaden moon
from the exercise ig $35 \mid a$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
since $\gcd(5, 7) = 1$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
\textbf{Exercise 4.} Let $a, b \in \mathbb{Z}$, not both zero, such that $5 \mid...
im thinking of something
as i said earlier, since $5 \mid a$ and $\gcd(a, b) = 7$ then $35 \mid a$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
let $a = 35k$ and $b = 7l$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
let $d$ be the $\gcd(2a^2 - b^2, a^2 - 3b^2)$ then $d \mid 2a^2 - b^2$ and $d \mid a^2 - 3b^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so you could see that $d \mid 5b^2$ and substitute the values
soft zealotBOT
1 divided by 0 equals Infinity
orchid coral
d divides 2(2a^2 - b^2) + (a^2 - 3b^2) = 5a^2 - 5b^2
So d divides 5a^2 too
gentle zephyr
i dont understand
leaden moon
where?
orchid coral
Basically, the HCF of two numbers divides both the numbers, and so it also divides any linear combination of those two numbers
gentle zephyr
I understood that 35 | a
gentle zephyr
Basically, the HCF of two numbers divides both the numbers, and so it also divid...
yeah
I understand that aswell
d = gcd(x,y) => d | x and d | y => d | ax + by
@leaden moon @orchid coral
orchid coral
i dont understand
what
leaden moon
gcd is defined as the highest common factor between 2 numbers
gentle zephyr
i dont understand how to solve the exercise
orchid coral
this part?
his statement of it clearly shows he understands that part
leaden moon
his statement of it clearly shows he understands that part
i guess so
gentle zephyr
help please
leaden moon
normally, to find the $\gcd$ with variables, you let the $\gcd$ as $d$ and try to eliminate a variable
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
d = gcd(2a^2 - b^2, a^2 - 3b^2)
d | 2a^2 - b^2
d | a^2 - 3b^2
yea
multiply by 2 on the bottom equation to eliminate a^2
gentle zephyr
also we know that
7 | a,
7 | b,
5 | a
so
35 | a
7 | b
leaden moon
yeah
because of that
we let a = 35k and b = 7l
with a note that gcd(k, l) = 1 because gcd(a, b) = 7
orchid coral
Subbing a = 35k and b = 7l, we get GCD(2a^2 - b^2, a^2 - 3b^2) = GCD(49(50k^2 - l^2),49(25k^2-3l^2) = 49 x GCD(50k^2 - l^2, 25k^2 - 3l^2), where k and l are coprime
Since k and l are coprime, so are k^2 and l^2
Replace k^2 and l^2 by p and q
We now find GCD(50p - q, 25p - 3q) where p and q are coprime numbers
soft zealotBOT
gentle zephyr
also you are overcomplicating it
gentle zephyr
Subbing a = 35k and b = 7l, we get GCD(2a^2 - b^2, a^2 - 3b^2) = GCD(49(50k^2 - ...
instead of using this, maybe we can usee that gcd divides both arguments
then get to something like d | a^2
gentle zephyr
what do we do after that
orchid coral
hm
leaden moon
case 1 is easy enough
so im not going to talk about that
case 2: d | a^2 and d | b^2
this is when you substitute a and b in
gentle zephyr
you mean a = 35k and b = 7q
thats what you want me to substitute
@orchid coral @leaden moon
leaden moon
yea
leaden moon
aight then
soft zealotBOT
leaden moon
yeh
gentle zephyr
yeah, right?
gentle zephyr
leaden moon
,rccw
soft zealotBOT
gentle zephyr
ok so basically I can get rid of the long gcd and just use this, d | 5a^2 and d | 5b^2
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
ok so basically I can get rid of the long gcd and just use this, d | 5a^2 and d ...
d = gcd(2a^2 - b^2, a^2 - 3b^2)
like there is no more use for this
I can just usee those two
d | 5a^2
d | 5b^2
@leaden moon @orchid coral
leaden moon
yea!
leaden moon
**1 divided by 0 equals Infinity**
and from that you get this
gentle zephyr
so either $d \mid 5$, or $d \mid a^2$ and $d \mid b^2$
we already know that d | 5
thats for sure..
leaden moon
how about $d \mid a^2$ and $d \mid b^2$?
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
we need to
scarlet sequoia
so either $d \mid 5$, or $d \mid a^2$ and $d \mid b^2$
Why so?
gentle zephyr
use a = 35k , b = 7q with gcd(k,q) = 1
scarlet sequoia
isn't 5 prime though?
But we're not in a 5 | ... situation
We're in a d | ...
And d isn't necessarily prime
leaden moon
But we're not in a 5 | ... situation
ill let you take over if that sounds good?
scarlet sequoia
Sry I can't I was just popping by
eternal matrix
?
final saddleBOT
?
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).
soft zealotBOT
gentle zephyr
@leaden moon @orchid coral
leaden moon
wait hold on
from $d \mid 5a^2$ and $d \mid 5b^2$ then $d$ must divide $\gcd(5a^2, 5b^2)$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
what?
leaden moon
you got $d \mid 5a^2$ and $d \mid 5b^2$ right?
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
ye
leaden moon
so $d$ actually divides $\gcd(5a^2, 5b^2)$, not 5
soft zealotBOT
1 divided by 0 equals Infinity
soft zealotBOT
gentle zephyr
@leaden moon @orchid coral
leaden moon
and $\gcd(a^2, b^2)$ is 49
soft zealotBOT
1 divided by 0 equals Infinity
final saddleBOT
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.
gentle zephyr
Is q^2 = l^2?
what?
orchid coral
Nvm that.
gentle zephyr
we are so close bro
leaden moon
yes!
leaden moon
since $\gcd(a, b)$ is 7
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
then it makes sense that $\gcd(a^2, b^2)$ is 49
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so substitute $\gcd(a^2, b^2) = 49$ into here
soft zealotBOT
1 divided by 0 equals Infinity
orchid coral
how do you know that gcd(25k^2, q^2) = 1
As 5 | a but (a,b) = 7, 5 does not divide b because if it did then (a,b) would have to divide 5.
So 5 does not divide b = 7q
So 5 does not divide q
So 5^2 does not divide q^2
So 25 and q^2 are coprime
So 25k^2 and q^2 are coprime because (25, q^2) = 1 and (k^2, q^2) = 1
gentle zephyr
As 5 | a but (a,b) = 7, 5 does not divide b because if it did then (a,b) would h...
im so fucked
gentle zephyr
then it makes sense that $\gcd(a^2, b^2)$ is 49
we dont know that gcd(a,b) is 7
soft zealotBOT
Result:
245
leaden moon
yea
leaden moon
so d in {5, 7, 35, 49}?
first of all\
soft zealotBOT
leaden moon
we can get rid of 5 and 35
because they gave $5 \mid a$, and $\gcd(a, b) = 7$ so definitely $5 \nmid b$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
because they gave $5 \mid a$, and $\gcd(a, b) = 7$ so definitely $5 \nmid b$
yeah
gentle zephyr
we can get rid of 5 and 35
how?
soft zealotBOT
1 divided by 0 equals Infinity
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
I dont understand
leaden moon
haiz
leaden moon
I dont understand
where?
leaden moon
**1 divided by 0 equals Infinity**
you get this?
gentle zephyr
no
but basically, if 5 does not divide b, then b does not have a 5 in his prime factorization
and if b does not have a 5 in his prime factorization, then neither b^2 which has the same prime factorization but squared
gentle zephyr
b = p1 . p2 . p3 . . .
b^2 = p1 ^2 . p2^2 . p3^2 . . .
leaden moon
and in b there's no 5 in it right?
gentle zephyr
ye
gentle zephyr
yeah
gentle zephyr
i mean (5^0)^2 = 5^(2.0) = 5^0 = 1
leaden moon
yeah
gentle zephyr
I dont understand your point
where are you going with this
only think I understand is that d | 245
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
or do you have to prove something like this?
leaden moon
yeah
gentle zephyr
then $5 \nmid 2a^2 - b^2$
we know that 5 | 2a^2 so 2a^2 = 0 (mod 5)
then 2a^2 - b^2 = -b^2 (mod 5)
leaden moon
assume $2a^2 - b^2 \equiv 0 \pmod{5}$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
and as you said, $2a^2 - b^2 \equiv -b^2 \pmod{5}$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so this leads to $-b^2 \equiv 0 \pmod{5}$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
and then -b^2 = 0 (mod 5) <=> 5 | b^2
gentle zephyr
yeah
leaden moon
so the only conclusion is that $5 \nmid 2a^2 - b^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
then this means $5 \nmid \gcd(2a^2 - b^2, a^2 - 3b^2)$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
which means $5 \nmid d$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
what?
5 | d
d | 2a^2 - b^2
d | a^2 - 3b^2
5 | d , d | 2a^2 - b^2 => 5 | 2a^2 - b^2
gentle zephyr
which means $5 \nmid d$
I dont get it
leaden moon
5 | d
d | 2a^2 - b^2
d | a^2 - 3b^2
5 | d , d | 2a^2 - b^2 => 5 | 2a^2 - b^2
but we just proved that $5 \nmid 2a^2 - b^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
<@&268886789983436800>
leaden moon
5 | d
d | 2a^2 - b^2
d | a^2 - 3b^2
5 | d , d | 2a^2 - b^2 => 5 | 2a^2 - b^2
so what you shown is a contradiction proof of $5 \nmid d$
soft zealotBOT
1 divided by 0 equals Infinity
final saddleBOT
@gentle zephyr Has your question been resolved?
gentle zephyr
so what you shown is a contradiction proof of $5 \nmid d$
that doesnt make any sense whatsoever to me
leaden moon
we proved that $5 \nmid 2a^2 - b^2$ right?
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
yes
leaden moon
you said that if $5 \mid d$, but because $d \mid 2a^2 - b^2$ then $5 \mid 2a^2 - b^2$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
you said that if $5 \mid d$, but because $d \mid 2a^2 - b^2$ then $5 \mid 2a^2 -...
what?
you are using the fact that divisibility is transitive
?
@leaden moon
are you here or not?
gentle zephyr
d | 245
leaden moon
**1 divided by 0 equals Infinity**
And this leads to a contradiction
Which means $5 \nmid d$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
,w prime factorization 245
soft zealotBOT
gentle zephyr
Which means $5 \nmid d$
yeah
leaden moon
And $d \mid 245$
gentle zephyr
5 | a => 5 does not divide d
otherwise
gcd(a,b) not 7
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
The factors of 245 are whatever you said
And after eliminating factors that 5 divides we get 7, 49
gentle zephyr
5 | d , d | 245 => 5 | 245
leaden moon
5 | d , d | 245 => 5 | 245
Yeah
leaden moon
**1 divided by 0 equals Infinity**
Do you agree with this?
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
Then $d$ can only be 1, 7, 49, 245, 5, 35
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
But we proved that $5 \nmid d$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
Values like $d = 5$ does not satisfy $5 \nmid d$
gentle zephyr
no
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
no
leaden moon
no wait no
?
leaden moon
5 | a => 5 does not divide b
Yeah
leaden moon
Assume $5 \mid b$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
And because $5 \mid a$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
Then $5 \mid \gcd(a, b)$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
but I understand that 5 does not divide b bro
Yea
gentle zephyr
yes I understand that 5 | a and 5 does not divide b
then I also know that d | 245
about d I dont know anything else
leaden moon
$d$ was like uhm
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
$\gcd(2a^2 - b^2, a^2 - 3b^2)$?
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
d = gcd(2a^2 - b^2, a^2 - 3b^2)
leaden moon
Yeah
leaden moon
Earlier we proved that $5 \nmid 2a^2 - b^2$
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
suppose 5 | d
5 | d , d | 2a^2 - b^2 => 5 | 2a^2 - b^2
we know that 5 | a and 5 does not divide b
2a^2 - b^2 = 0 (mod 5)
-b^2 = 0 (mod 5) <=> 5 | b
leaden moon
suppose 5 | d
5 | d , d | 2a^2 - b^2 => 5 | 2a^2 - b^2
we know that 5 | a and ...
Yep
That leads to a contradiction
gentle zephyr
ok we also know that 7 | a and 7 | b
wait I dont understand
245 = 5 * 7^2
we know d | 245
and now we know that 5 does not divide d
gentle zephyr
I dont get it
leaden moon
,w factors of 245
soft zealotBOT
gentle zephyr
245 = 5 x 7^2
leaden moon
d | 245
gentle zephyr
5 does not divide d
gentle zephyr
5 | d <=> d = 5k
leaden moon
5 | d <=> d = 5k
To eliminate all values that has this form
gentle zephyr
1 7 49 are the possible values of the gcd then
leaden moon
d = gcd(2a^2 - b^2, a^2 - 3b^2)
So basically $49 \mid 2a^2 - b^2$ and $49 \mid a^2 - 3b^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
So it can't be 1 and 7
gentle zephyr
7 | a <=> a = 7k
a^2 = (7k)^2 = 7^2 . k^2 let k^2 = q
a^2 = 7^2 .q
7^2 | a^2 <=> a^2 = 7^2.q
so basically d is 1 , 7, or 49
@leaden moon
leaden moon
So $a^2 = 49q$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
Similarly $b^2 = 49r$ for example
soft zealotBOT
1 divided by 0 equals Infinity
steady lark
should be 1 or 5
leaden moon
So $2a^2 - b^2 = 2 \cdot 49q - 49r = 49(2q - r)$
soft zealotBOT
1 divided by 0 equals Infinity
soft zealotBOT
1 divided by 0 equals Infinity
gentle zephyr
what are u talking about
steady lark
multiply the second term by 2 and subtract and same for the next from the original question
u should get 5a2 once and 5b2 once
so it is ofc 1 , and can be 5 as well
eh that kinda becomes a pattern though
steady lark
So it can't be 1 and 7
how come it wont be 1?
what was the original question , what did it ask ?
in english i mean
final saddleBOT
@gentle zephyr Has your question been resolved?
viral jewel
What is your question ❓
south dirge
its pinned
viral jewel
I think I get the solution.
We have:
[
\gcd(a,b)=7
]
This means there exist (m,n) such that:
[
a=7m,\quad b=7n,\quad \gcd(m,n)=1
]
Substituting into:
[
d=\gcd(2a^2-b^2,; a^2-3b^2)
]
we obtain:
[
d = 49 \gcd(2m^2-n^2,; m^2-3n^2) = 49d_1
]
Now:
[
d_1 \mid \left(2m^2 - n^2 - 2(m^2 - 3n^2)\right) = 5n^2
]
[
d_1 \mid \left(-3(2m^2 - n^2) + (m^2 - 3n^2)\right) = -5m^2
]
Hence:
[
d_1 \mid 5m^2
]
Therefore:
[
d_1 \mid 5\gcd(m^2,n^2)
]
Since (\gcd(m,n)=1), we have:
[
\gcd(m^2,n^2)=1
]
Thus:
[
d_1 \mid 5
]
So:
[
d_1 = 1 \text{ or } 5
]
If we take (m=2) and (n=3), this gives a value for (d_1).
Finally:
[
d = 49 \times 1 = 49
]
soft zealotBOT
William James Moriarty
final saddleBOT
@gentle zephyr Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
hollow radish
if a number is irrational in binary is it irrational in decimal and what if it's irrational in base 3 or even base 4 is even base 12 is it still irrational in decimal
final saddleBOT
if a number is irrational in binary is it irrational in decimal and what if it's...
blissful meadow
Irrationality isn't a feature of the representation of a number in a base, it's a feature of the number itself.
If your question is whether the number will be nonrepeating with infinitely many decimals, then yes.
hollow radish
I know but in base pi pi is simply 10 it terminates
blissful meadow
Yes well then by this argument any irrational number is "rational" in its own basis.
Again, irrationality is a particularity of a number, namely that it can't be expressed as a fraction of integers, not of how you represent it.
blissful meadow
Again, irrationality is a particularity of a number, namely that it can't be exp...
.
hollow radish
irrationality also means a number's digits won't repeat
i mean pattern wont repeat
blissful meadow
It's a consequence of the fact that you can't write irrationals as fractions that they have nonrepeating digits with no pattern.
hollow radish
If irrational in any natural base>2 it's irrational in base 10?
hollow radish
i mean 0.101001000100001..... (n)-> n (10) n still can't be represented as p/q where p,q∈ℤ
n∈ℕ n>1 True or false
<@&286206848099549185>
blissful meadow
!15m
final saddleBOT
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.
blissful meadow
And true, this is an irrational number.
final saddleBOT
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Available help channel!
wanton otter
do yall know which of the formulas wil appear in each number/section of the quiz? I'm kinda confused if all of them would appear in establishing identities and etc
final saddleBOT
do yall know which of the formulas wil appear in each number/section of the quiz...
sturdy flax
since they are all trig identities, most likely you're expected to use whichever that will fit the question as it is set?
wanton otter
since they are all trig identities, most likely you're expected to use whichever...
so ill just expect that any of the formuals could appear in each sections?
mossy inlet
probably, i mean, if you're comfortable using them it should be fine
can also derive most of them from each other if u forget (couhg cough me on my last quiz-)
sturdy flax
so ill just expect that any of the formuals could appear in each sections?
yes. I think the main takeaway here is that you should be prepared to use any of them in any question.
wanton otter
where would co function identities most likely appear since i havent really focused on that
wanton otter
i see
sturdy flax
where would co function identities most likely appear since i havent really focu...
the best way is to see if you have access to past year papers.
wanton otter
the best way is to see if you have access to past year papers.
he gave example problems should i share it to you?
sturdy flax
if you do, you can see which functions/identities appear more frequently in which sections.
rocky tusk
where would co function identities most likely appear since i havent really focu...
probably nowhere
wanton otter
oh
sturdy flax
that is, if you want to game the exam statistically. the best option is still to just learn it and expect anything anywhere, and use whatever necessary for the job.
rocky tusk
but it's pretty obvious if you remember a standard right triangle
wanton otter
that is, if you want to game the exam statistically. the best option is still to...
what's the best tactic with establishing identities
its genuinely confusing
rocky tusk
the sine of one acute angle is the same ratio as the cosine of the other acute angle and vice versa
from this you can get all of the cofunction identities
where the two acute angles are x and pi/2 - x
sturdy flax
to be quite honest, knowing the identities here is just half the battle when establishing identities.
the other half is algebraic manipulation.
and recognizing special forms like the difference of two squares (eg: 1 - cos^2).
sturdy badge
I get that its conditionally convergent, but why cant i use the divergence test to prove its not absolutely convergent? can I?
final saddleBOT
I get that its conditionally convergent, but why cant i use the divergence test ...
formal trail
if you were to apply the divergence test the sum of |a_n| you would find it to be inconclusive
sturdy badge
wouldnt the sum of 1/ln(6n) not be zero
formal trail
the divergence test asks about the limit of the terms
sturdy badge
yeah the limit is zero right?
formal trail
yes
final saddleBOT
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Available help channel!
solar lichen
final saddleBOT
solar lichen
Keep getting these wrong idk what I’m doing wrong ngl
rugged merlin
Keep getting these wrong idk what I’m doing wrong ngl
Show whar you have done
solar lichen
What was your logic
Not to sure to be honest
I was going to use this formula but now I realise that a isn’t actually 1/2
So I’m lost
stone wagon
this only works if it's of the form e^(ax+b)
but you have a square root there
and also it's divided by sqrt(x)
stone wagon
you need to do a u-substitution, letting u=sqrt(x)
solar lichen
you need to do a u-substitution, letting u=sqrt(x)
Can you demonstrate how to do that please
stone wagon
Can you demonstrate how to do that please
what's the derivative of sqrt(x)
solar lichen
0.5x^-0.5
stone wagon
cursed way to write it but sure
1/(2sqrt(x)) basically
you can actually use this
your f(x) = sqrt(x)
however note that f'(x) is 1/(2sqrt(x)) while you have 1/sqrt(x)
so you can write $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} = 2\int \frac{e^{\sqrt{x}}}{2\sqrt{x}}= 2 e^{\sqrt{x}} + C$
soft zealotBOT
artemetra
stone wagon
@solar lichen like that
solar lichen
Where on earth did 2 come in from though?
stone wagon
you multiply and divide it
they cancel
$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} = \int \frac{2e^{\sqrt{x}}}{2\sqrt{x}} = 2 \int \frac{e^{\sqrt{x}}}{2\sqrt{x}}$
soft zealotBOT
artemetra
stone wagon
Where on earth did 2 come in from though?
i wrote the middle step
final saddleBOT
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Available help channel!
mortal island
In this scenario….to which part does the 2x+6 multiply
final saddleBOT
In this scenario….to which part does the 2x+6 multiply
whole halo
In this scenario….to which part does the 2x+6 multiply
you arent done yet
next step, you multiply both sides by (x + 4)
with that, you get 8(x + 4) = 5(2x + 6) which you then simplify
mortal island
Slow down
I didn’t ask that-
I’m trying to get rid of the denominator on the LHS
So to which part of the other side do I multiply it
whole halo
youre free to process the information at your own pace
just because I type fast doesnt mean you have to read it fast too
just reread it again to know what to do
you can try it, too
mortal island
I type fast too but you’re multiple steps ahead of where I need to be
blissful meadow
So to which part of the other side do I multiply it
Like I said last night, there is no ambiguity as to "where" you multiply. You decided to multiply the LHS and the RHS by $2x+6$. This is a good move.
$$\frac{8}{2x+6}\cdot (2x+6) = \frac{5}{x+4} \cdot (2x+6) = \frac{5}{x+4} \cdot \frac{2x + 6}{1}.$$
soft zealotBOT
Azyrashacorki
blissful meadow
MY LORD AND SAVIOR HAS RETURNED 🙏🙏🙏🙏
Hahahah bless
blissful meadow
Yes, since that's how you multiply things :p
blissful meadow
We've all been there
compact laurel
Look I’m not the brightest and it’s early
OMG bro, just you wait.... lol
mortal island
Huh
compact laurel
Bro's a true mathematician
mortal island
I think I fucked up
mortal island
I meant just you wait.... the late nights will only continue, lol
Not after this semester is over
After this it’s prob and stats and im DONE
compact laurel
Not after this semester is over
No more math, eh?
compact laurel
I don’t intend to go into a math related field
Do you know what you wanna go into?
sturdy flax
,rccw
soft zealotBOT
mortal island
Do you know what you wanna go into?
Performing and theater
compact laurel
Performing and theater
That's sick bro!! Maybe you'll be required to sing the quadratic formula as an opera one day.... you never know...
mortal island
Hell nah
compact laurel
Hell nah
And for the record bro, "I, Me, Mine" has been stuck in my head ever since we talked yesterday, lol
mortal island
Peakkkk
compact laurel
I think I fucked up
What was the original question btw?
compact laurel
I think I fucked up
Just so I can help you figure out where you went wrong
mortal island
compact laurel
So, the first thing to always do in an algebra problem, is look for any restrictions (i.e., is there anything that x CANNOT equal?
mortal island
Well I already did it I just don’t think the answer looks right
silver maple
!show
final saddleBOT
Show your work, and if possible, explain where you are stuck.
mortal island
^^
silver maple
The working is quite messy. I can't read it
steel sonnet
Can U sing your working out pls?
mortal island
The working is quite messy. I can't read it
Mb 💀💀💀
compact laurel
Mb 💀💀💀
You wanna try doing it again?
Writing a bit neater?
If it's not too big of an ask
modest sequoia
your first step is wrong. you have done $\frac{x-1}{(x+1) \cdot x}=\frac{-6}{x} \cdot x$ to get rid of it.
soft zealotBOT
Krish
mortal island
Fuck
modest sequoia
you should be multiplying to the numerator
mortal island
FUCK
I always mess that up
My fatal flaw is multiplying anything with fractions
Does this look better (unfinished)
mortal island
Writing a bit neater?
My handwriting is not THAT bad
soft zealotBOT
mortal island
Idk how that couldn’t be read
compact laurel
My handwriting is not THAT bad
I didn't mean it as a sly 😭
mortal island
I know
mortal island
The other guy couldn’t
strange pelican
you wrote (x+1) multiplied in denminator on LHS , you should multiply in numerator
but you have seemed to cancel the denominator correctly
then you should use brackets, your final result is not correct
mortal island
There is no final result
It’s in progress
And I wrote it there because that’s where it fit and I don’t have to turn in this work
Unless my last step is wrong
modest sequoia
$-6(x+1) \neq -6x+1$
soft zealotBOT
Krish
modest sequoia
distribute to all terms
i highly recommend writing parentheses when multiplying anything with more than 1 term
strange pelican
Unless my last step is wrong
it is
mortal island
modest sequoia
you have written $-6 \cdot x + 1$, which is not the same
soft zealotBOT
Krish
mortal island
I just wanna be done
I don’t even have a passing grade on Ts yet and there are still 5 more sections
I’ve done 6 sections
modest sequoia
even if you consider them equal you wont reach the final
well yeah, beacuse they arent equal. after distributing properly its factorable after bringing everything to LHS
compact laurel
I don’t even have a passing grade on Ts yet and there are still 5 more sections
But that's at least a good thing. You still have 5 more sections. Try not to stress man. At the end of the day, a grade does not define you. You still have time buddy. I believe in you
mortal island
I don’t tho
I have 5 more assignments
All of which are the same length or longer
2 of which are proofs
And everything has to be done by Friday or I fail the class
compact laurel
And everything has to be done by Friday or I fail the class
Oh, it's all online?
compact laurel
I have 5 more assignments
Another recommendation is do an absurd amount of practice problems. Not to the point that you like kill yourself. But just do enough where you can do it without thinking, and it becomes second nature
For some, that's a harder spot to get to than others
mortal island
I don’t have time for that
compact laurel
I don’t have time for that
And know what, fair enough. And while it's easier said than done, try not to stress about it. Everyone I've met in this server is really nice. And everyone I've met would be willing to help you. So, this is judge free environment. You're not gonna be judged for, "not doing it right"
mortal island
Well good because I need more help
It’s hard to get into a groove when the type of problem changes every 2 questions
compact laurel
It’s hard to get into a groove when the type of problem changes every 2 question...
That's true, but you can also break things down, into categories, and always try to simplify your problem
For example, in this one, I see a hell lot of fractions
I don't like fractions
They ate me middle finger
mortal island
Yeah me neither
compact laurel
So, do you notice the common denominator on the 2nd, and third fraction?
mortal island
Yes
compact laurel
So, you can move the 2nd fraction to the other side of the equals sign, (changing the sign to a '-'), and then subtract the fractions
mortal island
Ahhhh
compact laurel
Lmk once you do that (if you're even doing this problem rn, lol)
compact laurel
So….7/x+5 = 1?
Cool. But now, you need common denominator. So, write one with a denom. of (x + 5)
mortal island
But 1/x+5 isn’t…..
compact laurel
You're correct, it isn't
mortal island
But that’s what you said to write
compact laurel
But when you're finding a common denom. you need to multiply both the numerator, and denominator by that factor to not change the value of the term
mortal island
1 with a denominator of x+5 is 1/x+5 😭😭😭😭
compact laurel
But when you're finding a common denom. you need to multiply both the numerator,...
So, I agree with you, 1/(x + 5) is not the same as 1. So to compensate, we also multiply the numerator by (x + 5)
compact laurel
So x-2/x+5 =0
How'd you get there?
mortal island
By doing the same thing you told me to do earlier
Subtract by 7/ x+5 on both sides
mortal island
Wdym
compact laurel
For what x does (x -2 )/(x +5) give you 0?
mortal island
It took the answer
compact laurel
So, then you ask yourself
Is 2 = to any of your restricted values (-5, and 5 in this case)
mortal island
I don’t have enough time to care about checking
compact laurel
Is 2 = to any of your restricted values (-5, and 5 in this case)
And since it's not a resitrcted value, you know it's an answer
compact laurel
We have no matching denominators
So, for this one, it's subtle, but you can force at least one common denom.
mortal island
Is it uhhhhh….the first one
compact laurel
(3 - x) is SOOO close to (x - 3). It's just off by a negative
So if you factor out a negative from (3 - x) -> -(x - 3)
compact laurel
Is it uhhhhh….the first one
Do you see that?
mortal island
Mueheheheh
final saddleBOT
Available help channel!
granite sage
Simplest way to solve it ?
final saddleBOT
Simplest way to solve it ?
cunning ridge
Simplest way to solve it ?
perhaps splitting and substitution?
vapid hound
Simplest way to solve it ?
do you know how to do u sub?
cunning ridge
but id say thats a long way
vapid hound
perhaps splitting and substitution?
i feel u sub and then using trig would be easiest
granite sage
do you know how to do u sub?
Yes
cunning ridge
i feel u sub and then using trig would be easiest
yeah
vapid hound
Yes
let u^2 be equal to the whole fraction
that would be how i’d approach it
because of the square root *
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Available help channel!
tall dagger
help
unreal robin
i have no idea how to solve this question
There's an algebraic translation of the intersection of two graphs (i.e the curve and the straight line)
tall dagger
umm
tall dagger
so u put the second equation in the place of y in the first equation?
south dirge
here you could sub the value of y from the linear eqn into the QE eqn
south dirge
so u put the second equation in the place of y in the first equation?
yup!
tall dagger
so im gonna find x then right?
south dirge
so im gonna find x then right?
yup!!
south dirge
And once you get x, you can put value of x into any eqn to get value of y!
tall dagger
ok
south dirge
Well eqn of x would give you two solns
tall dagger
yep
tall dagger
so how do i know whats x in A and whats the x in B ?
tall dagger
So you have two seperate points x coords
yep
south dirge
so how do i know whats x in A and whats the x in B ?
They can be anything!
tall dagger
and then i put those two x coordinates in the second equation one at a time and find two y ?
tall dagger
They can be anything!
oh ok
south dirge
Its not mentioned in the Qsn for A to be some specifc is it!
south dirge
and then i put those two x coordinates in the second equation one at a time and ...
Yup!
tall dagger
yea
south dirge
i got 1 more qeustion
sure!!
tall dagger
idk how to do question number 20 and 21
south dirge
idk how to do question number 20 and 21
.pin