#help-36
hidden birch
wraith crater
bro u suggested it
wraith crater
sorry i was talking to @hidden birch
soft zealotBOT
$Pure$
hidden birch
i think ur diagram is a ∗ bb ∗ which accepts any number of 'a's followed by at least one 'b' and then any number of 'b's thereafter
to make it a * b *
Keep state q0 as it is, with transitions on 'a' looping back to q0
add a trasitino from b from q0 to q1
in q1, have b loop back to q1 but do not allow trasition on a
sharp kindle
in q1, have b loop back to q1 but do not allow trasition on a
ahh, but i did the same in the diagram i drew the second time right?
the # means end state (same as double circle)
sharp kindle
ur diagram is correct imo
ok tysm, i got so confused since i thought my first one would also be correct... since its and infinite amount of a's with an infinite amount of b's but now that i see it, it can also lead to aba which shouldn't be possible in a*b*
sharp kindle
how do i close?
final saddleBOT
hidden birch
still shuttle
Hey! I'm hoping somebody can help me understand how to answer these questions. I'm basically unsure how to read a contour diagram. i know A through C and E are correct but I want to know why. Also, I don't know how to answer D for these reasons. tysm ;-;
final saddleBOT
hidden birch
@still shuttle do not occupy 2 channels
final saddleBOT
Available help channel!
compact ferry
final saddleBOT
compact ferry
i'm not sure what this question is asking of me
what does it mean the integral defining f(x)
tranquil pine
have you worked with and solved integrals before?
tranquil pine
all its asking is for you to solve the integral
compact ferry
do i use the first part of the fundamental theorem
tranquil pine
just use any part that gives you the answer in a way you understand the best
its not specifying which part
tranquil pine
yeah
compact ferry
this is the second part of the question how do i just find the limit of the equation i just got
bc isn't it just 1/inf which is 0
tranquil pine
this is the second part of the question how do i just find the limit of the equa...
u could graph the equation
final saddleBOT
Available help channel!
finite warren
I was wondering if anyone could help me with homework regarding electricity?
zinc geyser
post your quetion
final saddleBOT
I was wondering if anyone could help me with homework regarding electricity?
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
finite warren
post your quetion
How do I calculate currents?
finite warren
I have calculated that the parralel Ohm of r3 and r4 is 6
finite warren
where are you struggling
Ok so I calculated that the total Ohm is 24/10 Ohm
of R1, R2, R3 and R4
And I don't know how to continue now
final saddleBOT
Show your work, and if possible, explain where you are stuck.
shut gazelle
I have calculated that the parralel Ohm of r3 and r4 is 6
This is not right
finite warren
I don't think you guys will be able to understand this because of my handwriting but anything may help
shut gazelle
And if R34 (R3 and R4 combined) is 6/7
finite warren
I thought it was 42
shut gazelle
So your overall resistance is wrong
finite warren
Alright I recalculated it
its 164/69
((41/7).(4)/((41/7) + 4)
= 164/69
How do I continue?
shut gazelle
Find the overall current
Ohm's law
V = IR
You know V, and you need the total resistance
shut gazelle
No, you need to find the total resistance
shut gazelle
= 164/69
This is the resistance of when you combined R1, R2, R3, R4
You still have R5 and Ri
finite warren
You still have R5 and Ri
Oh, I thought that we excluded the first and last sets of resistance in our circuit
Alright
shut gazelle
Oh, I thought that we excluded the first and last sets of resistance in our circ...
You can't because it's still resistance in the circuit
finite warren
You can't because it's still resistance in the circuit
164/69 + 4 + 10 = 1130/69
And that should be the total resistance
shut gazelle
Now you can do V = IR to find the total current of the circuit
finite warren
Now you can do V = IR to find the total current of the circuit
And that is 93.5/(1130/69) = 5.71A
shut gazelle
Do you know your current and voltage rules for series and parallel circuits?
finite warren
Do you know your current and voltage rules for series and parallel circuits?
Not really, no
shut gazelle
Voltage in series drops, voltage in parallel stays the same
Current in series stays the same, current in parallel drops
Those are the rules you should know
finite warren
Right
shut gazelle
So you have a circuit that is Ri, R5, and the rest of the resistors simplified with R = 164/69, correct?
finite warren
So you have a circuit that is Ri, R5, and the rest of the resistors simplified w...
Yes
shut gazelle
Using those rules above, because those resistors are in series, what is the current through each one?
finite warren
Using those rules above, because those resistors are in series, what is the curr...
R1 currents stays the same, r2 the current stays the same, R3 and R4 the current drops
but the general current if you take all those resistors together stays the same since they're basically in series
shut gazelle
I'm not referring to that. I'm referring to the circuit that you have, with Ri, R5, and the R = 164/69
Those three resistances are in series
What is the current through each resistor
finite warren
What is the current through each resistor
Well, we have to calculate that
Ri is 5.71A, and R5 is also 5.71A
R = 164/69 is also 5.71A too isn't it
shut gazelle
Yes
finite warren
Alright
shut gazelle
Exactly
shut gazelle
Well, we have to calculate that
Ri is 5.71A, and R5 is also 5.71A
If you notice on the answer key, I5 = 5.71 A
finite warren
If you notice on the answer key, I5 = 5.71 A
Yes, and that's because all the current goes to a single resistor
shut gazelle
R = 164/69 is also 5.71A too isn't it
Use this, find the voltage through that resistor
finite warren
Use this, find the voltage through that resistor
Alright
V = IR
5.71 times 164/69 = 13.57V
shut gazelle
So that's the voltage through the 164/69 resistor
finite warren
Yes
shut gazelle
The 164/69 was originally a parallel circuit that was R2 and R = 41/7, correct?
finite warren
The 164/69 was originally a parallel circuit that was R2 and R = 41/7, correct?
Yes, that's correct
And you want to use V = IR to calculate them individually as well?
But the voltage is unique for each resistor so we can't do that
shut gazelle
No, recall the voltage and current rules
Because 164/69 has 13.57 V through it, and R2 and R = 41/7 were in parallel, that means voltage is the same, in each branch
So the R = 41/7 has 13.57 V, and R2 has 13.57 V
finite warren
so for R2, V = IR would be
I = V/R
13.57/4 = 3.39A
shut gazelle
Yes
finite warren
and for R = 41/7, it is 2.31A
shut gazelle
And that part of the circuit was R1 and R = 6/7
shut gazelle
Yes and first goes through R1, which means R1 = 2.31A
Yes
shut gazelle
Yes and first goes through R1, which means R1 = 2.31A
And the current in R = 6/7 is also 2.31 A
Because R1 and R = 6/7 are in series
finite warren
Yes
shut gazelle
So you know the current in R = 6/7, you can find the voltage through it
shut gazelle
Yes
finite warren
ooh
shut gazelle
So that's the voltage through R = 6/7
But it was in parallel, so what does that mean?
finite warren
and then 1.98/6 and 1.98/4 right
shut gazelle
Yes voltage stays the same
So you know voltage in R3 and R4 is 1.98 V so you can find current
finite warren
R4 = 1 ohm
ah damn I thought R4 was 4 ohm
finite warren
Close, R3 = 6 ohms
alright so 1.98/6 = 0.33A and 1.98/1 = 1.98A
shut gazelle
How do I calculate currents?
R4 is 1 ohm
shut gazelle
Not sure where you got 4 from
shut gazelle
You just need to know your parallel and series rules with current and voltage and apply those
final saddleBOT
thank you very much
If you are done with this channel, please mark your problem as solved by typing .close
finite warren
.close
night dragon
i am wrong or right
final saddleBOT
i am wrong or right
soft zealotBOT
Result:
226.19467105847
vital crag
show how you got 226.08
night dragon
soft zealotBOT
Result:
3.1415926535898
bronze compass
If we got a square and divide it by a straight g we get two parts with the same area. draw both diagonals and prove that g has to go trough the intersection of both diagonals
final saddleBOT
If we got a square and divide it by a straight g we get two parts with the same ...
bronze compass
I have tomorrow a math olympiad and want to train
What I would do in this exercise is first draw a square
In order for a square to be divided into two same areas, those areas have to bw a^2/2
Given that a is the length of one side
Thus, you can either cut one of the sides in half
Having a straight going trough the intersection point
Another way is to divide it into triangles with right angle with length a, with this we achieve it too
But how do I prove, that there are no more possibilities for dividing?
final saddleBOT
@bronze compass Has your question been resolved?
bronze compass
<@&286206848099549185>
digital steeple
if you draw a straight line through a square this line intersects the sides of the square. which cases do you see?
bronze compass
if you draw a straight line through a square this line intersects the sides of t...
What do you mean?
digital steeple
make a sketch.
bronze compass
make a sketch.
Or half one of the sides and make a rectangle
But how do I prove, that there are none anymore? Cause those 4 go trough the intersection of the diagonals
digital steeple
i see two possibilities (ok a third one is a special case):
the line intersects two opposite sides (right) or the line intersects two adjacent sides.
the third case (special one) is if the line is a diagonale.
do you agree?
bronze compass
But in those cases the areas are Not the same?
Oh no in your 2nd case they ate
Are
digital steeple
step by step. first: how can a line intersect a square.
do you agree with this two cases?
bronze compass
Ah no it does not intersect a square. A straight should part the square area in two even areas. And the statement is that all of those straights have the same point, which is the diagonal intersection point
bronze compass
do you agree with this two cases?
Yeah in this case the line intersects the square
Ah I understand
Either it cuts two opposite or two adjacend sides
So only such straights can part it
Now I need to limit them
digital steeple
exactly. lets start with the left part. the line intersects to adjacent sides. then on figure is a triangle.
right?
bronze compass
Yeah because the only way to divide an square into two areas with same area is by an triangle by a diagonal, as the we get a^2/2
digital steeple
exactly. and if the line is a diagonal it goes clearly through the intersection of the two diagonals, so, we are ready with this case.
bronze compass
Yes
digital steeple
now to the right part. which type are the two parts?
bronze compass
Well we need to part the area of the square into two even parts, which means either by cutting one of the sides by half, as per definition a^2/2 which means by rectangles, or by a trapez
digital steeple
both parts are trapezoids (a rectangle is a special trapezoid)
bronze compass
Ah yeah
With the rectangles they have to go trough the intersection point, as I cut the side in half and the intersection point is the middle point
But how do I prove it for the other trapezoids?
digital steeple
write the area of the two trapezoids as formula. and then use that they are equal.
bronze compass
1/2( x+y)*a =1/2 (a-x+a-y)*a
digital steeple
simplyfy.
bronze compass
a cant be zero so we can divide by a and 1/2
x+y = a-x + a-y
2x +2y = 2a
x+y = a
digital steeple
yes. write it as x = a-y. and then look at the sketch
bronze compass
Both sides of the two trapezoids are the same, but simply mirrored
digital steeple
exactly.
one trapezoid is a rotation with 180 degress of the other one. the only point (as center of the rotation) which can fulfill this, is ....?
bronze compass
The intersection point of the diagonal
digital steeple
well, now we are ready.
bronze compass
Alright thx!!
As I have proved that both cases go through the intersection point its proven
.close
final saddleBOT
final saddleBOT
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green plank
Any ideas as to what i did wrong?
final saddleBOT
Any ideas as to what i did wrong?
rose mountain
for part 1 your working out and answer don't match
I think you are in radians not degrees
green plank
I think you are in radians not degrees
Yeah your right i was in radians🤦♂️🤦♂️
I was so confused 😂
Ty
.close
!close
pulsar wadi
i dont know if im deriving with lhoptial wrong?
after deriving i got -2/x^2- 2/x+1
<@&286206848099549185> why dont the channels get sent to math help? Is something down in the server?
pulsar wadi
.close
tranquil pine
<@&286206848099549185> why dont the channels get sent to math help? Is something...
try to open a new one
pulsar wadi
try to open a new one
#help-7｜zen1thxyz yeah still not working there
final saddleBOT
<#423244559682764800> yeah still not working there
pulsar wadi
.close
final saddleBOT
languid pulsar
worldly vale
Bot broken gotta wait
final saddleBOT
Available help channel!
shut axle
using fundamental calculus theorem i need an f(x) so f(0)=0 and it´s derivative follow the intergal. i dont eaven know where to start
final saddleBOT
using fundamental calculus theorem i need an f(x) so f(0)=0 and it´s derivative ...
amber holly
According to the theorem, what is the integral on the left equal to? (Feel free to introduce a new function)
shut axle
it would be x*f´(√x-1)
amber holly
Are you sure that's what the theorem suggests?
shut axle
although i think the x^2+1 afect the result
shut axle
why F'(x) = xf'(sqrt(x-1))?
amber holly
Can you state the fundamental theorem of calculus?
shut axle
if i derivate the integral i need to take acount the x^2+1 of the limit and derivate as weel
amber holly
Right
amber holly
Yes, okay, let's use that version
shut axle
my f(t) it wuld be (t*f´(√x-1))
amber holly
Define F(x) = integral from 1 to x of t f'(sqrt(t-1)) dt so that F(x^2 + 1) will be the integral that we are considering
F'(x) would be xf'(sqrt(x-1)), right?
amber holly
Okay, and we are given that F(x^2 + 1) = 2x^2
Try differentiating both sides since we are able to express the derivative of F
shut axle
then (x^2 + 1) *f´(√(x^2 + 1) -1)=(x^2 + 1) *f´(√(x^2) = 2x^2?
amber holly
I think you forgot about chain rule
F(x^2 + 1) differentiated would be 2xF'(x^2 + 1) which is 2x(x^2 + 1)f'(sqrt(x^2+1-1))
And you forgot to differentiate 2x^2
shut axle
F'(x) would be xf'(sqrt(x-1)), right?
and why this then?
amber holly
So that I could go from 2xF'(x^2 + 1) to 2x(x^2 + 1)f'(sqrt(x^2+1-1))
shut axle
F´(x) would be 2xxf´(√x-1) = (2x^2)*f´(√x-1))
amber holly
Where did you get 2x from
shut axle
the chain rule, is the derivative from x^2 + 1
amber holly
You would have to use chain rule for F(x^2 + 1)
Not F(x)
And, again, recall how we defined F
shut axle
i am lose there.
if F(x) =∫(a,x) of tf´(√t-1) dt / F´(x) = xf´(√x-1)
and if
F(x) = ∫(a,x^2+1) of tf´(√t-1) dt / F´(x) = 2xx*f´(√x-1)
am i right?
amber holly
You are close
You just need to put x^2+1 instead of t when you are deriving the second equation
shut axle
in the top one?
shut axle
why?
amber holly
Because you should replace t with the upper bound of the integral in this case
And what's the upper bound in the integral that you mentioned in the second line?
shut axle
if F(x) = ∫(a,x^2+1) it is like a composite function
F(g(x)) g(x)=x^2+1
so F´(x) = F´(x) * g´(x)
amber holly
so F´(x) = F´(x) * g´(x)
Don't use the same letter for different things
You just ended up saying that g'(x) = 1
Anyway, yes, if F(x) = the integral from 1 to x and g(x) = x^2 + 1
Then the integral that we are considering would be F(g(x))
Whose derivative is F'(g(x)) * g'(x)
shut axle
perfect, i follow
amber holly
Whose derivative is F'(g(x)) * g'(x)
But that's the same as F'(x^2 + 1) * 2x
shut axle
F'(g(x)) = x*f´(√x-1)
amber holly
F'(g(x)) = x*f´(√x-1)
No, just F'(x)
shut axle
and g'(x)=2x
amber holly
But that's the same as F'(x^2 + 1) * 2x
We know that F'(x) = xf'(sqrt(x-1)), so F(x^2+1) = (x^2 + 1)f'(sqrt(x^2+1-1))
Hence the integral differentiated is 2x(x^2+1)f'(sqrt(x^2+1-1))
amber holly
We know that F'(x) = xf'(sqrt(x-1)), so F(x^2+1) = (x^2 + 1)f'(sqrt(x^2+1-1))
The first equation is due to the fundamental theorem of calculus and the fact that we defined F(x) to be the integral from 1 to x of t f'(sqrt(t-1)) dt
shut axle
let me se if i get it.
F(x) =∫(a,x) of tf´(√t-1) dt
F´(x) = xf´(√x-1)
F´(x^2+1) = (x^2+1)f´(√(x^2+1)-1)?
i dont see where the 2x come from
amber holly
Yes
amber holly
i dont see where the 2x come from
It appears when we multiply by the derivative of x^2 + 1
I think you keep confusing F'(x^2 + 1) with the derivative of F(x^2 + 1)
They are different things
The first is the derivative of F evaluated at x^2 + 1 and the second is the function F(x^2 + 1), whatever it may be, differentiated
shut axle
so F(x^2 + 1) = ∫(a,x^2+1) of tf´(√t-1) dt
and
F´(x^2+1) = (x^2+1)f´(√(x^2+1)-1)
amber holly
Yes
amber holly
I think you keep confusing F'(x^2 + 1) with the derivative of F(x^2 + 1)
So it's like the order of operations is different
To get F'(x^2 + 1) you first differentiate F(x) and replace x with x^2 + 1
To get the derivative of F(x^2 + 1) you first get the expression for F(x^2 + 1) by replacing x with x^2 + 1 in whatever F(x) is and only then you differentiate
Anyway, we get that (F(x^2+1))' = F'(x^2 + 1) * 2x = 2x(x^2+1)f'(sqrt(x^2+1-1))
shut axle
F'(x)= x* f(√x-1)?
amber holly
Yes
shut axle
now i evaluate that in x^2+1?
amber holly
Yes, and you will get F'(x^2 + 1)
shut axle
exellent, so
F´(x^2+1) = (x^2+1) * f´(√(x^2))
amber holly
Yes
shut axle
then F´(x^2+1) = (x^2+1) * f´(x)
amber holly
Would be better if you put |x| though I think
But I think that will be correct anyway
Anyway,
We were given that F(x^2 + 1) = 2x^2
So now 2x(x^2 + 1)f'(x) = 4x
shut axle
so 4x=(x^2+1) * f´(x)
amber holly
F(x^2 + 1) = 2x^2
(F(x^2 + 1))' = (2x^2)'
2xF'(x^2 + 1) = 4x
2x(x^2 + 1)f'(x) = 4x
shut axle
(F(x^2 + 1))' ≠ F´(x^2+1) ?
shut axle
then F´(x^2+1) = (x^2+1) * f´(x)
here
amber holly
Right
shut axle
and (F(x^2 + 1))' is like the derivative of a composite function?
amber holly
Yes
amber holly
Yes, due to chain rule
shut axle
so, 2x * (x^2 + 1) * f'(x) = 4x
shut axle
(x^2 + 1) * f'(x) = 2
amber holly
Right
amber holly
2x/(x^2 + 1)
amber holly
the f(x) woul de the integrate of that?
Correct
amber holly
2x/(x^2 + 1)
Put parenthesis when texting, I may have confused what you meant with $\frac2{x^2} + 1$ instead of $\frac2{x^2 + 1}$
soft zealotBOT
A Lonely Bean
shut axle
yes, sorry 2/(x^2 + 1)
amber holly
Do you know the antiderivative of 1/(x^2 + 1)?
shut axle
and to check if i am Right i evaluate the result in 0
amber holly
You will have the constant of integration and you should choose its value so that the function is 0 at x = 0, yes
shut axle
2/tan(x)
shut axle
yes
amber holly
But tan^-1(x) is not the same as 1/tan(x)
amber holly
tan^-1 and arctan are the same, neither of them are equal to 1/tan though
shut axle
yes, 1/tan = cotan
amber holly
Right, so, anyway, f(x) = 2arctan(x) + C
shut axle
and arctan (0) * 2 = 0
shut axle
c = 0 as weel
amber holly
Yes
amber holly
You are welcome
amber holly
It's fine, what matters is that now you've learned
amber holly
You too
shut axle
.close
midnight sparrow
How to show that gcd(a+b, lcm(a,b)) = gcd(a,b) with a and b strictly positive integers ? Does anyone have a hint or a method like I don't know what to do when I'm ask to show that something divide something else or to find gcd
final saddleBOT
How to show that gcd(a+b, lcm(a,b)) = gcd(a,b) with a and b strictly positive in...
midnight sparrow
arithmetic is a bit complicated for me
wraith crater
Show gcd(a+b, lcm(a,b) | gcd(a,b) and gcd(a,b) | gcd(a+b, lcm(a+b))
midnight sparrow
ye it's what I thought but don't know how to do that
wraith crater
Think about the `definition’ of gcd
midnight sparrow
just to be sure, gcd(a,b) is the unique integer d such that D(d) = D(a)nD(b), I'll try with this but all in arithmetic is about definition and I don't know how to use it correctly
I'll try it
wraith crater
gcd(a,b) divides a and b
any other Divisor of a and b divides gcd(a,b)
desert mantle
what is that D(n) notation?
midnight sparrow
what is that D(n) notation?
divisor of n
desert mantle
set of divisors?
desert mantle
(also that is false cause -d also works)
desert mantle
"is the unique integer d such that D(d) = D(a)nD(b)"
unique is false, because d and -d both satisfy that
midnight sparrow
yes but not the condition before
he need to satisfy both integer and that condition
so he is unique if he satisfy both
desert mantle
the word integer means that it can be both positive or negative
final saddleBOT
@midnight sparrow Has your question been resolved?
midnight sparrow
I'm stuck I don't know how to write this, I try to write it with the set of divisor but it gives me nothing
like is there a set of the divisor of lcm(a,b)
midnight sparrow
<@&286206848099549185>
final saddleBOT
@midnight sparrow Has your question been resolved?
midnight sparrow
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waxen prawn
Let $U \subseteq \mathbb{C}$ be an open set, $f_n : U \rightarrow \mathbb{C}$, $f : U \rightarrow \mathbb{C}$ be holomorphic functions and $z_0 \in U$ such that $f_n$ converges uniformly to $f$ over compact sets, $f$ is not identically zero and $f(z_0)=0$. Prove that there is some sequence $(z_n)$ and some $n_0 \in \mathbb{N}$ such that $z_n \rightarrow z_0$ and $f_n(z_n)=0 \ \forall n \geq n_0$.
final saddleBOT
Let $U \subseteq \mathbb{C}$ be an open set, $f_n : U \rightarrow \mathbb{C}$, $...
soft zealotBOT
Casiel368
final saddleBOT
@waxen prawn Has your question been resolved?
waxen prawn
<@&286206848099549185>
final saddleBOT
@waxen prawn Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
turbid carbon
How would I solve the equation X^2 + X = A^2 +A by completing a square
final saddleBOT
How would I solve the equation X^2 + X = A^2 +A by completing a square
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
loud sundial
Note you want to make the left hand side a perfect square
how familiar are with you with completing the square
turbid carbon
im familar with completing a square but the extra variable is confusing me
loud sundial
Don't worry about that for now
just focus on the left hand side
what do you need to do to make it a perfect square
turbid carbon
+1/4
but then i would add +1/4 on the other side and the equation would be (x-1/2)^2 = (a-1/2)^2
loud sundial
but then i would add +1/4 on the other side and the equation would be (x-1/2)^2 ...
it should be plus instead of minus
$x^2-x+\frac{1}{4}$ would be $\left(x-\frac{1}{2} \right)^2$
soft zealotBOT
Civil Service Pigeon
loud sundial
but then i would add +1/4 on the other side and the equation would be (x-1/2)^2 ...
but yeah this is fine
turbid carbon
and then im lost from there
loud sundial
Well we need to solve for x, but it's 'stuck' under a square
so what do you thnk we can do
turbid carbon
ig square both sides
turbid carbon
oh i meant to root both sides
turbid carbon
ill get X-1/2 = A-1/2
soft zealotBOT
Civil Service Pigeon
turbid carbon
+- X ???
soft zealotBOT
Civil Service Pigeon
turbid carbon
so ig i just move -1/2 from here right?
turbid carbon
so the answer should be X = A, -A + 1
loud sundial
so the answer should be X = A, -A + 1
close
I think you made a sign error with your second solution
oh wait
💀
I carried on the error I said not to make
loud sundial
**Civil Service Pigeon**
this error
it should be $$x+\frac{1}{2}=\pm \left(a+\frac{1}{2} \right)$$
soft zealotBOT
Civil Service Pigeon
final saddleBOT
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Available help channel!
reef canyon
final saddleBOT
reef canyon
can I use the fundamental theorem of calculus then integrate that derivative twice?
reef canyon
cause ftoc1 gives you the derivative
if you integrate it once you go back to the original function
if you integrate again you get the integral of the original function which is what we want
atleast that's what Im thinking
hybrid heath
Try u substitution
reef canyon
substitute what?
unique aspen
?
you want to find integral from 4 to 7 of this function g(x) dx
by FTC this is equal to G(7) - G(4) where G is an antiderivative of g
so you want to integrate g once to get G
and then find G(7) and G(4)
final saddleBOT
@reef canyon Has your question been resolved?
hard mantle
Substitute u = f(x) and du = f’(x) dx
final saddleBOT
final saddleBOT
Available help channel!
jolly vector
final saddleBOT
jolly vector
i have to find the positive and negative intervals
and the answers r saying its all negative
but hows that possible
how is it decreasing
from negative ♾️ to positive ♾️
shut gazelle
Can you send the full image for more context?
shut gazelle
Then send the full problem that it's asking you to do
jolly vector
Then send the full problem that it's asking you to do
isnt it increasing from - ♾️ to 7/3
warm ether
i think youre confusing positive with increasing
warm ether
increasing would mean the gradient is positive
jolly vector
the graphs increasing
jolly vector
how
soft zealotBOT
warm ether
youre doing ii?
jolly vector
was it asking for
the linear?
ouh ya
help
ouhemgee
yes
im just confusing myself at this point
😭🙏
thank u
.close
final saddleBOT
final saddleBOT
Available help channel!
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robust mulch
!show please
final saddleBOT
Show your work, and if possible, explain where you are stuck.
twilit oar
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Available help channel!
chrome gust
final saddleBOT
chrome gust
how was this changed
craggy thunder
e^ln(3x) = e^6
craggy thunder
yes
chrome gust
so log_e(3x)=6
craggy thunder
yes
craggy thunder
craggy thunder
basically
normal locust
hello
elder garden
can anyone help with where the forces are?
and whether my arrows are correct
oops sorry
normal locust
what is the formula, for discovering the cathetus
elder garden
I'll move away now sorry!!
normal locust
pls
final saddleBOT
@normal locust Has your question been resolved?
final saddleBOT
@normal locust Has your question been resolved?
red leaf
is chem math fine to ask?
final saddleBOT
is chem math fine to ask?
red leaf
like is it 2.50 or 2.27?
iron mist
like is it 2.50 or 2.27?
Show your work
red leaf
Show your work
,calc 2.38*294
soft zealotBOT
Result:
699.72
,calc 2.50*280
soft zealotBOT
Result:
700
Recheck your ideal gas law
red leaf
Recheck your ideal gas law
ik but the gas laws need to equal
if i do boyles law i get 2.27
but that does not equal to the same number
NVRM
im actually so dumb
bro
so sorry
.close
wintry cedar
How do I do this? (This is Simplifying Trig Identities)
final saddleBOT
How do I do this? (This is Simplifying Trig Identities)
zinc geyser
How do I do this? (This is Simplifying Trig Identities)
damn too bright
is it $\int \frac{sin^2 a}{\tan^2 a} dx$?
soft zealotBOT
Bettim
soft zealotBOT
Sp00ked
wintry cedar
How do I do the division?
zinc geyser
$1 - \frac{sin^2a}{\tan^2a}$
soft zealotBOT
Bettim
zinc geyser
?
wintry cedar
yes.
zinc geyser
take the second term
expand the tan
$1 - \frac{\sin^2 a}{\frac{sin^2 a}{\cos^2 a}}$
soft zealotBOT
Bettim
wintry cedar
the sin would cancel each other out
soft zealotBOT
Bettim
wintry cedar
sin^2.
zinc geyser
lol
wintry cedar
😆
wintry cedar
We haven't even touched intergration.
zinc geyser
okay bcz in the image you sent i thought it was integration sign
it wasnt clear haha
wintry cedar
My bad.
I was trying to make the one clear.
I have one more if that's fine with you.
zinc geyser
yea go on
wintry cedar
zinc geyser
you forgot the cos next to tan^2
so its
$\cos a + \cos a \frac{sin^2 a}{\cos^2 a}$
soft zealotBOT
Bettim
zinc geyser
now can you cancel any terms in the right term?
wintry cedar
the cosine?
wintry cedar
so it would be cos a + sin^2a/cosa right?
wintry cedar
to remove the denom. right?
zinc geyser
you should get like $\frac{\cos^2 a}{\cos a} + \frac{\sin^2 a}{\cos a}$
soft zealotBOT
Bettim
zinc geyser
to remove the denom. right?
yes
wintry cedar
1/cosa?
wintry cedar
sec
wintry cedar
I'm a little confused
wintry cedar
now multiply and divide the first term by cos a
on this.
zinc geyser
on this.
we have addition righr?
basic rule of addition of fractios is that
bases must be equal right
thts what we are doing
making the bases equal
wintry cedar
so it would be cos a + sin^2a/cosa right?
I basically wrote this
I assumed we multi the base of the fraction to get the sin^2 alone.
cosA/1?
@zinc geyser ?
zinc geyser
I basically wrote this
yea now uhhh....
see
$4 + \frac{4}{3}$
how would you do this?
soft zealotBOT
Bettim
wintry cedar
Still slightly confused.
wintry cedar
Wouldn't you multi the right side by cos/cos too?
soft zealotBOT
Bettim
wintry cedar
**Bettim**
I thought we were multiplying both sides by cos/cos
Why not both the left and right sides?
zinc geyser
I thought we were multiplying both sides by cos/cos
nah just the first term
wintry cedar
Why?
zinc geyser
Why not both the left and right sides?
sinse the second term already has cos in its base
we dont bother
wintry cedar
We want to get like terms?
wintry cedar
okay
Now I understand.
I've got another.
If we remove the bases,
it would just be sin^2(x) - cos^2(x) ?
zinc geyser
$a^4 - b^4 = (a^2 -b^2)(a^2 + b^2)$
soft zealotBOT
Bettim
zinc geyser
use this on your numerator
wintry cedar
it would just be sin^2(x) - cos^2(x) ?
this?
soft zealotBOT
Bettim
wintry cedar
hold on.
zinc geyser
from here its pretty easy
wintry cedar
those two cancel out.
zinc geyser
yes
wintry cedar
sin^2 + cos^2 = 1
unique aspen
can someone explain to me what exactly the steps im supposed to be taking here im kinda lost and confused on what im being asked to do
final saddleBOT
can someone explain to me what exactly the steps im supposed to be taking here i...
shut gazelle
No, the equation is r = sin theta
You have theta from the tables, use that to find r
So r = sin(0) for the first one
Then continue
unique aspen
isnt r =sin(0) = 0
unique aspen
1/2
shut gazelle
Continue with the rest of the table
lapis ridge
Do you guys need help?
shut gazelle
Do you guys need help?
Please don't open channels asking to look for help. Just help where you can
.close
final saddleBOT
shut gazelle
Do you guys need help?
#help-5 message
You don't need to ask this either. If you can help, then provide help, and if you can't help, then don't help or leave useless comments
final saddleBOT
Available help channel!
wintry kindle
very brief one as I haven't used W-Lambert before, how do you solve
x*a^x = a for x?
final saddleBOT
very brief one as I haven't used W-Lambert before, how do you solve
x*a^x = a fo...
@wintry kindle Has your question been resolved?
mint orbit
so
im trying to remember how to do these
we at least have to change the base
I think using $a^x = e^{x\log a}$ gets you 90% of the way there?
soft zealotBOT
jan Niku
mint orbit
you can use a substitution, if it makes you more comfortable
but I'd just stare at $$x e^{x \log a} = a$$ and make a mental wishlist
soft zealotBOT
jan Niku
mint orbit
does that help? @wintry kindle
wintry kindle
does that help? <@1000186270078750740>
hm can't I multiply by log(a)
soft zealotBOT
jan Niku
wintry kindle
hm does that solution hold if I instead have x*a^x >= a @mint orbit
or do I need to then take a different branch other than W0
mint orbit
oh man now youre really gonna test my memory
theres only two branches for real ... a here
real x?
to the wikipedia page 
wintry kindle
a and x are real in my case btw :)
mint orbit
i should just point you to the wiki lol
wintry kindle
ok
mint orbit
wintry kindle
the dumb thing is I utilise this for an analysis task I failed for like 6hrs so out of desperation I now have a long solution that's completely off-topic
mint orbit
so, its the sign on your x, i guess
wintry kindle
yes x will be positive, so W0(x) then ig
wet escarp
What were the fourier series formulas derived from?
final saddleBOT
What were the fourier series formulas derived from?
wet escarp
the formulas for a0, an, bn
warm forge
Which formulas?
wet escarp
warm forge
They are convolutions, integrals.
wet escarp
I understand how to use them but I'm having trouble understanding where they come from / how they are derived
warm forge
If you pose the question the other way round and then solve for "a0" you get this answer.
wet escarp
If you pose the question the other way round and then solve for "a0" you get thi...
what exatcly do you mean?
warm forge
So, pretent you don't have a fourmula for a0, then write down what it should do. Then solve.
warm forge
warm forge
A0 ... the integral of f(x)*cos(0) - same as for a0
wet escarp
I think I understand. THank you!
final saddleBOT
@wet escarp Has your question been resolved?
azure basin
Can someone tell me where I went wrong? (definitely a lot wrong here i think). approximating area using riemann sums taking the left endpoint of the rectangle slices where you make 5 equal rectangle slices
final saddleBOT
Can someone tell me where I went wrong? (definitely a lot wrong here i think). a...
azure basin
the last line is completely wrong but before that i had -1/3 💀
tulip coyote
Well for one, you're supposed to be approximating rather than actually taking the limit
azure basin
the way my prof taught us was to evaluate the summation and take the limit of that as n approaches infinity
tulip coyote
Furthermore, you want the left endpoints, but this choice and the summation start don't seem to agree (you'd want 0, 1, 2, 3, 4, but you have 1, 2, 3, 4, 5)
azure basin
ah ok
cyan hearth
is that a mixture of discrete math and calculus ?
tulip coyote
the way my prof taught us was to evaluate the summation and take the limit of th...
Difference for this one is that you're choosing n=5
azure basin
is that a mixture of discrete math and calculus ?
no this is intro to integral calculus, my notation si probably just weir
tulip coyote
If you wanted to work it out fully, you'd want to take the limit sure, but that gets you the exact answer rather than an approximation
azure basin
this is what i was taught 😭 ive previously learned how to evaluate integrals like this in high school but not in this stupid difficult summation way
cyan hearth
riemmannnnnnnnnnnnnn
azure basin
riemann 😔
cyan hearth
hate that man (?
modest birch
@vital crag
tulip coyote
(if you put n=5 in here, you'd get the right Riemann sum approximation)
cyan hearth
had*
azure basin
huh wdym
azure basin
as in i substitue n=5 into that equation?
tulip coyote
as in i substitue n=5 into that equation?
If you did, that would give you
approximating area using riemann sums taking the right endpoint of the rectangle slices where you make 5 equal rectangle slices
azure basin
ah ok
tulip coyote
Note how I changed it to right
azure basin
youre right about how evaluating the limit gives you the exact answer, my prof did say thats how you evaluate for the exact and not approximating
so if i wanted to just approximate then i would just sub n=5 right there and not bother with finding the limit?
okk
how would i change it to taking the left endpoint instead?
would i just do i-1?
okk
tulip coyote
would i just do i-1?
How I've usually seen it recommended!
azure basin
and then if i wanted to take the middle engpoint then i-0.5 ?
tulip coyote
Believe that works 
azure basin
omg this is making so mch sense 
also just checking but the amount that you subtract is just the length of one rectangle right
so like if i were to split this integral into only 2 slices i would do i-2.5?
for the left endpoint
cyan hearth
tulip coyote
so like if i were to split this integral into only 2 slices i would do i-2.5?
As in you wanted to do this but instead of 5, you'd want 2 slices of equal length?
azure basin
As in you wanted to do this but instead of 5, you'd want 2 slices of equal lengt...
yes
tulip coyote
No problem 
azure basin
.close
ivory vessel
$\log_2{(x-\sqrt{x^2-1})}\log_3{(x+\sqrt{x^2-1})} = \log_6{|x-\sqrt{x^2-1}|}$
soft zealotBOT
FungusDesu
ivory vessel
i dont know where to start since i cant multiply 2 log together can i
ivory vessel
Raise 6 to the power of both sides
$6^{\log_2{(x-\sqrt{x^2-1})}\log_3{(x+\sqrt{x^2-1})}} = 6^{\log_6{|x-\sqrt{x^2-1}|}}$
soft zealotBOT
FungusDesu
vital crag
That's what I meant yea. But I dunno if it'll work
final saddleBOT
@ivory vessel Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
stray stump
This limit evaluate to infinity right?
final saddleBOT
This limit evaluate to infinity right?
urban sky
This limit evaluate to infinity right?
Yes
final saddleBOT
final saddleBOT
Available help channel!
zinc stone
anyone want to explain Accurately solve using routine/non-routine operations the approximate percentage error in the time-period calculation if your measurement of length is 3% high and G is measured 2% too small.
final saddleBOT
anyone want to explain Accurately solve using routine/non-routine operations the...
@zinc stone Has your question been resolved?
zinc stone
no
final saddleBOT
@zinc stone Has your question been resolved?
jolly vector
final saddleBOT
soft zealotBOT
jolly vector
i got a
but for b
i duno what to equal them all to
how do i find the total # of time
it takes them all
they gave me the time two people take
im guessing its something to do with both
viral oriole
@jolly vector how did you do a?
final saddleBOT
@jolly vector Has your question been resolved?
waxen prawn
Find all Möbius transformations $T$ that verify simultaneously:\
$T(\hat{\mathbb{R}}) = \hat{\mathbb{I}},\ T(\hat{\mathbb{I}})={z \in \mathbb{C} / |z|=1},\ T({z \in \mathbb{C} / |z|=1})=\hat{\mathbb{R}}$.
final saddleBOT
Find all Möbius transformations $T$ that verify simultaneously:\\
$T(\hat{\mathb...
soft zealotBOT
Casiel368
waxen prawn
My plan is to find every transformation that satisfies each condition individually, and then finding the intersection
I got the first part already, but the other two got tough quickly
I have a weird idea though. If I manage to get the second condition right, I should be able to translate that into the third condition easily, but following the inverse path. So instead of considering the transformations that map the real axis to the unit circle, I consider all of their inverses. It might not be a nice thing to work with, but I guess it's worth a try.