let me think
#help-23
rapid blade
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@jaunty trench Has your question been resolved?
jaunty trench
that just affects precision
Ik but in every election there’s always 2 places after decimal
opal tartan
which thing?
thin bridge
that's their desired precision
opal tartan
i think so
if you are dealing with experimental values, you have to take care of the significant figures as well
thin bridge
its precise enough, but not that ridiculous
they have the data to get values with more precision and choose not to display it
jaunty trench
its precise enough, but not that ridiculous
But is 31,9% precise?
jaunty trench
What’s sf and dp?
thin bridge
short for significant figures
and decimal places
safe radishBOT
@jaunty trench Has your question been resolved?
jaunty trench
short for significant figures
and decimal places
What’s significant figures?
safe radishBOT
What’s significant figures?
jaunty trench
ITS the same? @thin bridge
thin bridge
what's the same?
jaunty trench
Significant and decimal places?
thin bridge
no, they're not the same
safe radishBOT
@jaunty trench Has your question been resolved?
jaunty trench
no, they're not the same
Explain
thin bridge
they're two different ways to measure precision
rounding something to a certain number of decimal places concerns digits after the decimal point
where rounding to a certain number of sig figs, you'd consider that amount of overall digits
jaunty trench
I don’t understand.
thin bridge
look up rounding decimal places
and rounding significant figures
spice wing
This may help
safe radishBOT
@jaunty trench Has your question been resolved?
pastel lake
❗️❗️❗️
topaz tree
Seems like your calculator doesn't support imaginary numbers
sqrt( 16 -20 ) = sqrt(-4)
which is not real
pastel lake
How do I do it in my calculator?
From where did u get 20
Also 16 it’s not in the question
thin bridge
they got -20 in the vid because they goofed
it's wrong
16-20 come from simplifying the stuff under the radical
what's 4^2
and 4(1)(5)
thin bridge
you can use your calc to simplify the stuff under the root for you if you wanr
pastel lake
How can u do that?
pastel lake
Like I do 4^2 then under it 4(1)(5) ?
pastel lake
Wait I’ll try
thin bridge
literally just enter in that "stuff"
and this isn't really something you should be resorting to your calc
thin bridge
yes
pastel lake
But it should give me root-20
nimble abyss
i think whats happened here is the quadratic has only complex solutions, and your calculator says error as the value under root is -ve, so unless complex solutions isnt covered in the class, i dont think she made an error, it just has to be solved partially by hand
nimble abyss
i think for this specific calculator model you have to set it to a mode that can use complex numbers
pastel lake
Okay wait
nimble abyss
if i remember correctly, click shift then setup and look for cmplx
pastel lake
I’ll use complex mode
nimble abyss
yeah and then reput in the expression
nimble abyss
did you put only the root?
pastel lake
I put the same thing in the complex mode
nimble abyss
puut the whole expression on the screen, not just the root and tell me what you get
pastel lake
But she wants us to do step by step
pastel lake
She wants us to do step by step in the exam
Like simplify each of them
Not all of them
thin bridge
sqrt(-4) will give 2i
nimble abyss
first simplify terms under the root
4^2-4(1)(5)
16-20
=-4
thin bridge
your calc is fine
nimble abyss
first simplify terms under the root
4^2-4(1)(5)
16-20
=-4
then seperate it into root(-1) and root(4)
root4 is 2
so it is 2i
pastel lake
first simplify terms under the root
4^2-4(1)(5)
16-20
=-4
I’ll do like this in the exam if it’s wrong idc it’s her mistake
nimble abyss
then seperate it into root(-1) and root(4)
root4 is 2
so it is 2i
(-4+2i)/2
divide all by 2
pastel lake
I’ll do -4
nimble abyss
you get -2+i
nimble abyss
I’ll do like this in the exam if it’s wrong idc it’s her mistake
im assuming thats what she wants
nimble abyss
you get -2+i
since it was originally a root -2-i is also a solution
does this solve your problem?
pastel lake
Thanks for the help
nimble abyss
np
safe radishBOT
@pastel lake Has your question been resolved?
solemn shell
Help me with this problem please.
safe radishBOT
Help me with this problem please.
solemn shell
I keep getting 8/7 or 1 1/7
chrome stag
just like multiply it out
solemn shell
algebra calculator online gave me this answer
paper idol
ig u expand it?
deft hull
Yes, 14 should be the answer.
solemn shell
i dont get how
deft hull
Use (a + b)^2 = a^2 + 2ab + b^2 to expand both the terms.
First term will come out to be 1 / 7+4*rt(3)
Second term will come out to be 1 / 7-4*rt(3)
paper idol
then rationalize the fraction
deft hull
then rationalize the fraction
Yup :).
solemn shell
Use (a + b)^2 = a^2 + 2ab + b^2 to expand both the terms.
how do u use that to expand it
deft hull
(1/2+rt(3))^2 = 1 / (2+rt(3))^2 = 1 / (4 + 3 + 22rt(3)) = 1 / (7+4*rt(3))
You can expand the second term similarly.
paper idol
$(1/2+sqrt(3))^2 = 1 / (2+sqrt(3))^2 = 1 / (4 + 3 + 22sqrt(3)) = 1 / (7+4*sqrt(3))$
flat frigateBOT
calinn
solemn shell
why cant it be
split ether
(a + b)^2 ≠ a^2 + b^2
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dapper lake
can someone explain functions for me
safe radishBOT
can someone explain functions for me
safe radishBOT
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spiral saddle
At $2!:!48$ what is the degree measure of the smaller angle formed by the hour and minute hands of a $12$-hour clock?
safe radishBOT
At $2\!:\!48$ what is the degree measure of the smaller angle formed by the hour...
flat frigateBOT
938c2cc0dcc05f2b68c4287040cfcf71
nimble abyss
At $2\!:\!48$ what is the degree measure of the smaller angle formed by the hour...
the minute hand is 12 mins away from the 0 position and that is a fifth of the minutes, so that angle is 72
the 2 is a sixth of the way past, so 360/6=60
and the hour hand moves 48/60% more than this
so 4/5x30=24
add them all up and you get 156
safe radishBOT
the minute hand is 12 mins away from the 0 position and that is a fifth of the m...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
nimble abyss
sure give me a few mins
spiral saddle
I will wait
nimble abyss
do u wannt me to write down the angle values or do u wanna work through it yourself?
spiral saddle
do a drawing
nimble abyss
with the angle values written or without?
spiral saddle
everything
as long as its self explanatory
otherwise dont include
how do I solve this?
nimble abyss
the minute hand is 12 mins away from the 0 position and that is a fifth of the m...
.
if theres anything u dont understand tell me
spiral saddle
eine moment bitte
nimble abyss
take your time
nimble abyss
the arc from 2 to 3 is a twelfth of the circle, and the length the hand moved is 4/5
so multiply all three
another way to think of it is it moved 4/5 the path along a 30 degree sector
so 4/5x30
nimble abyss
what does that even mean bruh
sorry for the bad wording
flat frigateBOT
Result:
24
Result:
156
156 degrees
spiral saddle
.close
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buoyant coral
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junior smelt
Given $T(t) = \frac{24}{T_1 e^{-kt} + 1}$, they tell you effectively that $T(1) = 4$ and $T(5) = 8$
flat frigateBOT
@junior smelt
buoyant coral
yes but what do I do with that
cuz if I plug in 4=24/(1e^-kt + 1)
i dont get ln 2.5/4
@junior smelt
junior smelt
You create simeltaneous equations to find T1 and k
junior smelt
cuz if I plug in 4=24/(1e^-kt + 1)
You put t = 1 here, which from there, rearrange
Then do the same with t = 5 and T = 8
buoyant coral
can you show me @junior smelt
junior smelt
Show you putting in t = 5 and T = 8, or rearranging the first?
junior smelt
Note that $T_1$ is not 1, as you wrote, you're supposed to put $T = 4$ and $t= 1$ into $T(t) = \frac{24}{T_1 e^{-kt} + 1}$ to get
[
4 = \frac{24}{T_1 e^{-k} + 1}
]
You can then multiply and divide to turn that into
[
T_1 e^{-k} + 1 = \frac{24}4
]
I shall let you try to do the same for $t = 5, T = 8$
flat frigateBOT
@junior smelt
junior smelt
(also I'm sure finding what 24/4 works out to isn't too challenging-)
junior smelt
T1 isn't 5, it's this
buoyant coral
yeah but how
junior smelt
In particular, 5e^k
junior smelt
<@788085606483361802>
and e.g. rearrange the equation to get $T_1 e^{-k} = 6 - 1 = 5$ thence $T_1 = 5e^k$
flat frigateBOT
@junior smelt
junior smelt
this one
junior smelt
Or alt, as e^{-k} = 1/e^k so then multiplying both sides by e^k
safe radishBOT
@buoyant coral Has your question been resolved?
daring cedar
how do i solve this system of EQ $\begin{cases} cosx+cos(x+y)=0 \ cosy+cos(x+y)=0 \end{cases}$
flat frigateBOT
swisher
tardy mango
It's trivial to see that $\cos x =\cos y$
flat frigateBOT
Civil Service Pigeon
daring cedar
where do i go from here
daring cedar
that gives cosycosx - sinysinx
tardy mango
I meant these
daring cedar
wym?
tardy mango
$\cos x +\cos (x+y)=2 \cos \left(\frac{x+2y}{2} \right) \cos \left(\frac{-y}{2} \right)$
flat frigateBOT
Civil Service Pigeon
daring cedar
oh i thought we were ignoring the first cosx sorry
daring cedar
**Civil Service Pigeon**
how does this help though
daring cedar
so one answer is y= +-pi but what about cos((x+2y)/2)
tardy mango
so one answer is y= +-pi
this part is wrong
what's the general solution for cos x = 0?
actually I should backtrack
did you solve cos x = cos y yet
daring cedar
no
tardy mango
do that first
it makes the rest a lot easier
since you'll have it all in terms of one variable
daring cedar
x= arccos(cosy) ?
tardy mango
I meant the general solution
daring cedar
x=y ?
tardy mango
what abt the periodicity?
daring cedar
+kpi
tardy mango
the period of cos is 2pi but lemme just step back again
yk how cos is an even function, right?
this means that along with x = y, x = -y also works
but you need to account for the periodicity
so $\cos x=\cos y \implies x=y+2k\pi, x=-y+2k\pi, k \in \mathbb{Z}$
flat frigateBOT
Civil Service Pigeon
daring cedar
ah okay
tardy mango
now put that into, say, $\cos x+\cos(x+y)=0$
just forget the sum to product for now since the (x+y) as an entity was confusing
flat frigateBOT
Civil Service Pigeon
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@daring cedar Has your question been resolved?
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bitter steppe
safe radishBOT
bitter steppe
this isnt a dot product is it
median vigil
it is the dot product
frozen veldt
that's only if a and b are the canonical basis vectors
i.e. (1, 0) and (0, 1)
often denoted i and j
if a and b are arbitrary vectors then you can use bilinearity
which basically means to treat the dot product like ordinary multiplication and FOIL (expand) out the parenthesis
safe radishBOT
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bronze yoke
I don't get how to solve this 
safe radishBOT
I don't get how to solve this <:bleak:877093280905592842>
bronze yoke
This is what I've tried so far
frozen veldt
,w (1/(2^x-1))+1/(4^x-1)
bronze yoke
woah what's that
frozen veldt
yeah there might be some trick to it but idk
since it's injective you could guess the soution and show that it's injective and thus the solution is unique
only problem is that no real solution seems to exist
bronze yoke
yeah there might be some trick to it but idk
photomath seems to solve it pretty easily
bronze yoke
x=2 according to photomath which seems about right
flat frigateBOT
frozen veldt
multiply both sides by (t-1)(t^2-1) and it becomes a cubic equation
and if you can guess one of the solutions then you can factor it into a second degree polynomial equation
but by this point it's pretty clear that this is not an ordinary homework exercise
this is some competition math problem or some crazy extra assignment
frozen veldt
your teacher is sadistic I'm not gonna lie
bronze yoke
damn
frozen veldt
but yeah there could be some simple trick to it but who knows
Photomath's solution is alright but of course solving a third degree polynomial equation is always a bit of a pain in the ass
nah what am I saying
bronze yoke
a third degree polynomial equation?
frozen veldt
that becomes a pretty easy equation once you simplify t^2-1 into (t+1)(t-1)
and then multiply both sides by t-1
it's actually a pretty good problem
I applaud your teacher
bronze yoke
I don't cus the assignment is due tomorrow
frozen veldt
it becomes a mere second degree equation
bronze yoke
and this is only the second sub question in the second question out of 4 questions in the assignment
frozen veldt
,w solve for t in 1+1/(t+1)=2/5 * (t-1)
I should probably not be doing math at this hour 😂
bronze yoke
lmao
flat frigateBOT
frozen veldt
yeah definitely a quadratic equation
A classic difference of squares
should've known
bronze yoke
and then multiply both sides by t-1
like this?:
(1/(t-1) + 1/(t+1)(t-1))*t-1 = (2/5)*t-1
frozen veldt
yee exactly
bronze yoke
doing this is so tough after just having an intense 3 hour study at the uni
frozen veldt
I mean it's a clever little problem once you figure it out but it's not like you're learning any super generalizable skills here beyond "always be on the lookout for a difference of squares"
which also is only really important for some contrived school exercises
bronze yoke
good to hear that because I'm probably not gonna see that in the future too
damn multiplying the whole thing by t-1 is pain
frozen veldt
Like basically the moral of the problem that I didn't spot originally was that you can turn 1/(4^x - 1) into 1/((2^x)^2 - 1) which is a difference of squares and becomes 1/( (2^x + 1)(2^x - 1) ) and then you can multiply both sides by 2^x - 1 and then you get a problem that you can solve with the quadratic formula
bronze yoke
I mean it's a clever little problem once you figure it out but it's not like you...
they do these a lot in the assignments I noticed
there's almost always some hidden binomial formula in some of these
frozen veldt
yeah it's always the same few algebraic identities recycled over and over in different settings
if it's a weird simplification problem then it's usually (a-b)(a+b)=a^2-b^2, sometimes (a+b)^2=a^2+2ab+b^2 can come up, and on very rare occasion you might see the mythical a^3-b^3 = (a-b)(a^2+ab+b^2)
but that's it basically
and then exponent rules etc.
frozen veldt
yeah it's like a funny trick
I think there might've been a general formula for a^n - b^n
bronze yoke
so eventually I get to ((t+1)+(t-1))/(t-1)(t+1) right? @frozen veldt
frozen veldt
this should be one of your intermediate equations
bronze yoke
uh
frozen veldt
after that you just multiply by t+1 and collect like terms and expand out some parenthesis
bronze yoke
frozen veldt
,w simplify 1+(1/(t+1))-(2/5)*(t-1)
flat frigateBOT
frozen veldt
,rotate ^(lmao what am I saying with the above query)
flat frigateBOT
frozen veldt
,w simplify (1+(1/(t+1))-(2/5)*(t-1)) * (t+1)
flat frigateBOT
frozen veldt
this is probably what your polynomial should look like
bronze yoke
uh I'd rather focus on getting this first
frozen veldt
How do you have (t-1)^2(t+1) in the denominator
like you've somehow managed to combine the fractions I guess
which is probably okay
bronze yoke
yeah I combined the fractions so that I can then multiply it by t-1
actually it does make more sense to just
wait I'll redo that part
frozen veldt
yeah idk if you combined the fractions correctly
being tired, and having to scroll through endless messages to see the original equation and trying to make sense of someone else's handwriting renders me pretty useless here tbh 😂
bronze yoke
being tired, and having to scroll through endless messages to see the original e...
you should be proud of yourself for being able to decipher my enchantment table writing
you helped me quite a bit too
frozen veldt
haha I'm glad
bronze yoke
Ok I successfully reached this
frozen veldt
congratulations
frozen veldt
your next step is probably gonna be this
,w (1+(1/(t+1))-(2/5)*(t-1)) * (t+1) real roots
flat frigateBOT
frozen veldt
yeah that still makes sense
bronze yoke
damn I'm so glad I went full out on my first assignments when they were still relatively simple
frozen veldt
I mean Photomath did a lot of the heavy lifting here tbh
I'm actually kinda impressed because I always just use wolfram alpha
bronze yoke
I mean Photomath did a lot of the heavy lifting here tbh
yeah it just skipped so far I had no idea how it got there
why do you multiply the whole thing by t+1?
frozen veldt
I moved everything to the lefthand side to get something=0 and then I got rid of the annoying denominator
alternatively you could combine the fractions first and then multiply by the denominator
but it's not like I was gonna start combining any fractions at this hour
bronze yoke
oh ok
I have to write why I do every step so for this one I'll write "multiply by (t+1) to get rid of the annoying denominator"
hehe
frozen veldt
hmm yeah I guess it doesn't even matter what order you do things in
you can multiply by that first and THEN move everything to the left hand side
either way works
but yeah, very based and math pilled
bronze yoke
I'll draw a little angry emoji next to it too
frozen veldt
that's cute
I mean, combing fractions is a good thing to do when you're not sure what else to do and it will always help you out
but sometimes you can take little shortcuts where you multiply by the denominator directly and then just remember to multiply every single term
which is easy to forget
when it comes to solving equations it's like painting, there's not a right or a wrong way to do it
except technically speaking you can do it wrong which is what usually happens to me
bronze yoke
if math is painting then my math would look like this
wait for (-2/5)(t-1)(t+1) do I use the binomial formula for (t-1)(t+1) or just do go from left to right and multiply ((-2/5)*(t-1))*(t+1)
I think the second one makes more sense
frozen veldt
if math is painting then my math would look like this
Let me just say real quick that this picture speaks to me on a spiritual level
frozen veldt
I think the second one makes more sense
you can do either one, but remember that it's not the binomial formula, it's the difference of squares formula
so (t-1)(t+1) actually becomes t^2 - 1
of course it's a fun exercise to expand out the parenthesis manually and notice that you get the same thing
bronze yoke
so (t-1)(t+1) actually becomes t^2 - 1
I always google "binomial formulas" for this
good to know the name thanks
frozen veldt
Well in combinatorics there's this thing called the binomial formula which is actually like a big complicated formula and it gives the coefficients of the nth power of any binomial i.e. (a+b)^n
so I guess (a+b)^2 is not a binomial formula but its coefficients are given by the binomial formula so that's why I got confused
but of course (a-b) is a binomial the same way as (a+b) is so yeah 😂
frozen veldt
I always google "binomial formulas" for this
I guess it would probbaly make sense that these would be called "binomial identities" or something like that
But "binomial formula" is something different and it's related to (a+b)^2
I think my native language might actually call those binomial formulas I don't know
it's been forever since I was in high school
yeah I don't really know what I'm talking about tbh
bronze yoke
I did something wrong ugh
I got (-2/5)t^2+t+12/5 which leads to x1 = 6/5, x2 = -16/5 which is not right
bronze yoke
I did it again and I'm getting the same result huh
I'll try to explain what I did
so from this
I got t+1 + 1 -2/5(t-1)(t+1) = 0
then I got t + 2 - ((2/5)*((t^2)-1)) so t + 2 - (2/5)t^2 + 2/5
hm
which is then t + 12/5 -(2/5)t^2
frozen veldt
seems good so far
bronze yoke
-(2/5)t^2 + t + 12/5
bronze yoke
perhaps
probably
I'll try again
ah I see where I made a mistake in the quadratic formula
frozen veldt
sqrt(1+2*2*2*2*3/(5*5))
bronze yoke
I forgot to divide by 2a
bronze yoke
then there's me who forgot to divide by 2 in the first place LMAO
bronze yoke
finally I reached t1 = -3/2, t2 = 4
I'll take a well deserved break
I think I know where to go from here
thank you so much for the help and I also appreciate the other general advice, that's broadened my knowledge @frozen veldt I'm really thankful
frozen veldt
thank you so much for the help and I also appreciate the other general advice, t...
aww that's super kind of you to say, enjoy your break and good luck with your future math problems! I'll also take a good night's sleep
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analog lake
this is a question from an sat mock exam i took recently:
"A parabola with the equation y = x^2 - 2x + 16 is graphed in the xy-plane. A line with the equation y = -6x - k, where k is a constant, intesects the parabola at exactly one point. What is the x-coordinate at the point of intersection?
A. -12
B. -2
C. 2
D. 12"
i answered with choice A (i guessed between A and B because i wasnt sure how to solve it) but the answer is B. can someone please explain how to get to the answer of -2?
gusty trench
the two functions intersect when they equal each other
this becomes a question of solving x^2 - 2x + 16 = -6x - k knowing that there's only one point of intersection
that becomes x^2 + 4x + 16 + k = 0
now if a quadratic only has one solution, you know its determinant must be 0
determinant is b^2 - 4ac where b = 4, a = 1, and c = 16 + k
rustic goblet
now if a quadratic only has one solution, you know its determinant must be 0
discriminant*
analog lake
oh that makes sense now
rustic goblet
let's not invoke the determinant here 
analog lake
solve for k
gusty trench
and then you can solve the quadratic for x
rustic goblet
gusty trench
yes
then you just solve the quadratic, which you already know only has the one solution
rustic goblet
gusty trench
yes
analog lake
makes sense
analog lake
really?
gusty trench
there's a way to solve it without solving for k
your parabola is an upward facing parabola
when a quadratic has two solutions, the parabola intersects the x axis twice, so in the case of this upward opening parabola the vertex would have to be below the x axis
when a quadratic has no solutions, the parabola is fully above the x axis and the vertex is above the x axis
but when a quadratic has one solution, it only intersects the x axis once, precisely at the vertex
royal kiln
are you able to use calculus?
analog lake
nope i havent gotten into calculus yet
gusty trench
if you can write x^2 + 4x + 16 + k = 0 into vertex form, you can just solve for the x coordinate of the vertex without needing to solve for k
that x coordinate is exactly your one solution
analog lake
i dont see why you'd need calculus..?
gusty trench
if you can write x^2 + 4x + 16 + k = 0 into vertex form, you can just solve for ...
and notice if you try writing this in vertex form by completing the square it'll look like (x+2)^2 + ... = 0
analog lake
if you can write x^2 + 4x + 16 + k = 0 into vertex form, you can just solve for ...
oh yeah that makes sense
hmm
since this is for the sat, which method of solving do you think would be faster
royal kiln
this is a question from an sat mock exam i took recently:
"A parabola with the ...
-12 is the value of k
analog lake
yeah neil just went over that with me
royal kiln
determinant is b^2 - 4ac where b = 4, a = 1, and c = 16 + k
that was clever
gusty trench
thank you 
there is a slightly faster way to solve this using calculus but that's overkill for the sat
useful if you know the trick ofc and in future math classes you might need to be able to solve a problem like this using the calculus approach and not the algebra approach, but this way should make the most sense
somber cape
Does this seem right?
safe radishBOT
Does this seem right?
hollow thunder
stoic dune
The division step is weird. (x - y)(x^2n +... ) = 0 implies that one of the two factors are 0.
lean otter
I need help with probabilities guys😭
stoic dune
All the trolls showing up
stoic dune
somber cape
The division step is weird. (x - y)(x^2n +... ) = 0 implies that one of the two ...
hmm i’m not sure i’m following
i got it from this
stoic dune
Let's say I want to solve:
(x - 1)(x - 2) = 0
I don't divide by x - 2.
I might be being pendantic. Your solution is good otherwise
It's just odd to write that part with a division
Actually, how do you know that other factor is always positive?
somber cape
Because the first and last term will alaways be positive since its a square and for any other combination of signs we will get an alternating sequences for the middle terms
if x and y are positive it will all be positive tho
I think i’m just relying on the fact that the power will decrease by 1 each time for x and increase for y
So the powers will always have the same parity
meaning that when they are even it will be a power of two or a square
so it must be alternating
and relying on the fact that the first and last terms are positive means that it has to be positive
I think thats correct but correct me if you see any flaws in that
willow pulsar
https://docs.google.com/spreadsheets/d/1ZJxRnDyN-l4CtfAQk9Cx1Z5I4UPWY1uFbgoMHvV2S1k/edit?usp=sharing I need help figuring out a growth formula 😄 Given a breeding cooldown in Minecraft and a rate of growth from baby to adult for sheep, I need to figure out how to take a sample of starting sheep and figure out the amount of sheep given any time. I can't figure out how to factor in NEW sheep then joining the mating pool
safe radishBOT
https://docs.google.com/spreadsheets/d/1ZJxRnDyN-l4CtfAQk9Cx1Z5I4UPWY1uFbgoMHvV2...
willow pulsar
The end goal is to figure out, given time, how much total wool could be extracted given a starting population of sheep that are sheared and bred maximally
So we have a time it takes for a sheared sheep to regrow its coat and a probability of that happening within that time
We also have how long the breeding cooldown is for the two sheep that breed
dull sequoia
I think a simulation might be easiest for this
willow pulsar
I think a simulation might be easiest for this
How would I put that together
dull sequoia
Just code up what you have
willow pulsar
idk how to do that lmfao
dull sequoia
An analytic solution seems a bit complicated
willow pulsar
Wouldn't I still need a formula
solar hazel
idk how the minecraft mechanics work but the population function probably satisfies a difference equation
(similar to a differential equation if you are familiar with that, but discrete)
willow pulsar
idk how the minecraft mechanics work but the population function probably satisf...
Basically the sheep can breed every five minutes. It takes 20 minutes for a baby to mature
Then it can breed
dull sequoia
Oh is it not random
dull sequoia
Oh that’s a lot simplier
You should be able to find the DE that describes the population over time
dull sequoia
You might be a bit out of luck here
willow pulsar
That's why I came here
Is there a way I can just make the spreadsheet do the math for me
steel marten
I don't think calc 1 has diff equations right
solar hazel
you can simulate it but probably need some basic coding at least for that
dull sequoia
Let’s use 5min blocks then, every time step, with a starting adult population of x, you get x adults and x/2 babies
Except, once you’re past the 3rd time step
You start adding the first group of babies into the adult pool
willow pulsar
Well 1st time step is x/2 babies. Second is X babies
dull sequoia
Let’s say x(t) is the number of adults at time t
Then we have the conditions that x(0) = a,
x(1) = a,
x(2) = a,
x(3) = a,
x(4) = x(3) + 0.5x(0),
x(5) = x(4) + 0.5x(1),
…
x(t) = x(t-1) + 0.5x(t-4)
willow pulsar
Series 😭 (or sequence idk)
dull sequoia
And then I think you do
$\lambda^{t+4} - \lambda^{t+3} - 0.5\lambda^{t} = 0$
flat frigateBOT
Frosst
dull sequoia
This gives you $\lambda^4-\lambda^3- 0.5 = 0$
flat frigateBOT
Frosst
dull sequoia
,w x^4-x^3-0.5 = 0
flat frigateBOT
willow pulsar
Oh you factored out lamda^t?
dull sequoia
Yeah I can’t quite remember entirely how this works but it’s how you solve difference equations
willow pulsar
I am a little lost how you got here
$\lambda^{t+4} - \lambda^{t+3} - 0.5\lambda^{t} = 0$
flat frigateBOT
GargatheOro
dull sequoia
But I’m not sure what the general solution to 4th degree ones are like
willow pulsar
,w x^4-x^3-0.5 = 0
Also what does x actually represent in this case
I am attempting to make my spreadsheet just crunch this using the interval approach
dull sequoia
So it’s $x(t) = A(-0.669)^t + B(1.253)^t + (0.2-0.744i)^t+(0.2+0.744i)^t$
flat frigateBOT
Frosst
dull sequoia
Then we have the conditions that x(0) = a,
x(1) = a,
x(2) = a,
x(3) = a,
x(4...
You can just use the last line then
willow pulsar
I need to figure out how to make sheets sum all intervals together 😄
dull sequoia
You need to give it x(0) through x(3)
willow pulsar
Oh I guess it wouldn't sum them
But I need some way for it to work in sequence
Sequences and series were always my weakest area
dull sequoia
You can just do in the A column 0 1 2 3 4, …
That’s t
In the B column you put a a a a
Then =Cell above + 0.5* cell 4 above
willow pulsar
True
dull sequoia
So say in B5 you put =B4 + 0.5*B1
Then drag it down
Now the B column is the shearable population at time t
willow pulsar
Is there some general equation I could use to not have to manually enter all that out? Like 20,000 intervals for example. Or a way I can find one analytically
dull sequoia
**Frosst**
It’s a bit ugly
willow pulsar
What do A and B represent
dull sequoia
You put this back into the recurrence relation then solve for the 4 constants
So it’s $x(t) = A(-0.669)^t + B(1.253)^t + C(0.2-0.744i)^t+D(0.2+0.744i)^t$
flat frigateBOT
Frosst
dull sequoia
The numbers you currently see control how fast it grows
The ABCD are the initial conditions
So you need to solve for them
If you haven’t studied DEs before it’s gonna be pretty foreign
But that’s the power of DEs, we can talk about systems that grow at a rate dependent on the current state
willow pulsar
Well damn now I wanna learn DE's
I see what the hype is
I study biology, I can think of so many applications for them
dull sequoia
DEs are like the baby version of what you actually use
Since they are univariatr
Most modelling stuff are PDEs, partial because they depend on different variables differently
Diffusion, waves, etc
willow pulsar
god damn
tf I somehow end up with decimals after 40 minutes. How can I have half a sheep 
Oh wait, that's cuz one sheep didn't have a mate in a scenario. Which means I need to round down every interval
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mossy ridge
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mossy ridge
When they expand it somehow else they get different critical points
idk how to approach these type of problems to solve them concretely
Because expanding it somewhat else shouldnt give a diff result 😭
safe radishBOT
@mossy ridge Has your question been resolved?
mossy ridge
i on expanding got sin³2x-sin²2x-sin 2x+ 1
Differentiating that i get points such that cos 2x= 0 or sin 2x=1
safe radishBOT
@mossy ridge Has your question been resolved?
safe radishBOT
@mossy ridge Has your question been resolved?
flat jolt
Low batch mai ho kya?
spice wing
1+sin2ø + (1-sin2ø)²
= 2 - sin2ø + sin²2ø
f'(ø) = 2sin4ø - 2cos2ø = 0
or sin4ø = cos2ø
or cos2ø(2sin2ø-1) = 0
if cos2ø = 0
Then , 2ø = pi/2
Or ø = pi/4
Check 2nd derivative with thia value^
If 2sin2ø = 1
Or 2ø = pi/6
Or ø = pi/12
spice wing
Or ø = pi/12
Also here solve for other values of ø in [0,2pi] and then check them too
Then add
stone cedar
Hello guys
safe radishBOT
Hello guys
stone cedar
This question was in my exam today
Im pretty sure its 45° but what is the best way to solve it
I reached a point where cosx = sinx which means the angle is 45°
tardy mango
$\sin x \cos x-\cos^2 x=0$
flat frigateBOT
Civil Service Pigeon
tardy mango
Factor out the cos x and run from there
(This does mean your answer isn’t fully correct)
safe radishBOT
@stone cedar Has your question been resolved?
zealous pike
safe radishBOT
zealous pike
for #25, the answer is d right?
main mural
yes
tardy mango
Recognise them as the definition of the derivative (or use the trig addition formulas)
zealous pike
my idea was to just ignore all of that limit shit and get the derivative
tardy mango
Yeah you’re better off interpreting the limits as derivatives
zealous pike
so like for 40, i rewrite it as sec^2(pi/6)
and then i just figure out that angle and then think about about what the side lengths would be?
and then from there i just use tan = opposite / adjacent
is that right or am i a silly billy
tardy mango
so like for 40, i rewrite it as sec^2(pi/6)
This is fine
Calculate the trig function however you want to
zealous pike
this exam would be so much easier if i was allowed to use a calculator lmao
@tardy mango is the answer for 40 4/3?
tardy mango
<@560122149126275093> is the answer for 40 4/3?
tardy mango
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Available help channel!
lean otter
A numbers digits' sum adds up to 45. It is the smallest natural number with this property.
If 2 of its digits switch places with each other, the natural number n will form, being the smallest number bigger than m
safe radishBOT
A numbers digits' sum adds up to 45. It is the smallest natural number with this...
lean otter
A numbers digits' sum adds up to 45. It is the smallest natural number with this...
The digits are different 2 by 2, so it can be abab but not aabb
Don't leave me please
why do people do this pls give me a sign
oak tangle
the smallest such number is 99999 but no digits can be exchanged here so you would need a 6 digit number
lean otter
but the digits have to be different 2 by 2
so it can be 12121212 for example
but not 112221
oak tangle
it can only have 2 different digits?
oak tangle
and the same digits cant be together?
lean otter
yes
oak tangle
for a 6 digit number of the form ababab, 787878 might work
lean otter
but the smallest natural number bigger than that is 787870
and if you switch any of the digits you can't get that
oak tangle
doesn't it have to be the smallest natural number with the same property?
lean otter
doesn't specify
but this would result in the last 2 digits being ab, where ba is ab+1
so 10a +b would be 10b+a+1
Ø
oak tangle
can you send the complete instructions given for the task as it is?
lean otter
no
it's for my final grade
basically I have 15 of these very difficult exercises this being one of them and I'm getting graded on 2 random ones
and this could be one of them
I can never know
oak tangle
thats.. a weird grading system
lean otter
it's not
that's how she decided to give our final grade
there are exercises with different difficulties
and these are the highest difficulty ones cuz yk I want a good grade
this is the only one idk
so
Anyway I'll close the channel
.close
safe radishBOT
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Available help channel!
golden badge
.
safe radishBOT
safe radishBOT
@lean otter Has your question been resolved?
pliant inlet
I dont know where to even begin here
safe radishBOT
I dont know where to even begin here
pliant inlet
is it (n-3)! * nC3?
safe radishBOT
@pliant inlet Has your question been resolved?
safe radishBOT
@pliant inlet Has your question been resolved?
safe radishBOT
@pliant inlet Has your question been resolved?
pliant inlet
is it (n-3)! * nC3?
pretty sure this solution is missing something
This is about counting labeled trees
covert yoke
@pliant inlet imagine you have a very large number of nodes, if only 3 are leaf nodes then you have only two branching points. These branching points can either be at the same location (1->3) or in two different locations. In the first case, this turns into a partitioning problem, where, if the branching point is at k nodes from the root, and there are n nodes total, how many ways can you partition n-k into three non-zero parts. This can be found with stars and bars. So it's a sum from k=0 to n-3 of the stars and bars problem.
For the second case now there's four pieces to worry about beneath the node at k. I haven't worked out an elegant way to attack this, but maybe this will put you on the right track
covert yoke
I doubt it
Oh, of course, the second case is also just stars and bars.
Also it depends on how you count trees, but either way there's some double counting you need to handle
Or rather either some under counting or some over counting
So if by your definition of tree the following are equivalent:
°
| \
° °
|
°
And its mirror image
°
| \
° °
|
°
Then you've over counted, and you need to use inclusion and exclusion to fix.
Otherwise you've under counted with the second case because you need to consider left then right branches, vs left then left, and right then right, and right then left.
Hth
pliant inlet
nC3 * (n-3)^n-5
but idk if this considers
but we're looking at the node structure of leaves and internal nodes
we need to restrict to just internal nodes
bleh
safe radishBOT
@pliant inlet Has your question been resolved?
pliant inlet
So if by your definition of tree the following are equivalent:
```
°
| \
° °
|...
mirror images are counted as different trees
for example
the second and fourth trees are mirrors and counted as diff trees
and the label is 1,2,3,4 starting from the bottom right and going clock-wise
sturdy pilot
for example
where are the root nodes in this image?
sturdy pilot
for n nodes and 1 leaf it should be 1 tree
for n nodes and 2 leaves it should be n - 2 trees
sturdy pilot
I dont know where to even begin here
Let $f_k(n)$ be the number of trees with n nodes and k leaves. Then,
$f_1(n) = 1 \
f_2(n) = n - 2 \
f_3(n) = \sum_{i = 1}^{n - 2} f_2(n - i) = \sum_{i = 1}^{n - 2} n - i - 2 \
\vdots \
f_k(n) = \sum_{i = 1}^{n - k + 1} f_{k - 1}(n - i)$
flat frigateBOT
Existentialistic
sturdy pilot
$\sum_{i = 1}^{n - 2} n - i - 2 = \sum_{i = 1}^{n - 2} n - \sum_{i = 1}^{n - 2} i - \sum_{i = 1}^{n - 2} 2 = n(n - 2) - \frac{(n - 2)(n - 1)}{2} - 2(n - 2) = n^2 - 2n - \frac{n^2}{2} + \frac{3n}{2} - 1 - 2n + 4 = \frac{n^2}{2} - \frac{5n}{2} + 3$
flat frigateBOT
Existentialistic
sturdy pilot
The idea is there are n - 2 possible lengths for the third branch, so we just add up all the possible 2-leaf trees with the remaining nodes.
I think I might be misunderstanding the question though, because google is giving me a different answer.
ah, I'm forgetting that forks don't need to occur at the root
safe radishBOT
@pliant inlet Has your question been resolved?
tidal thorn
The order for transformation matrix multiplication is Scale, Rotate, Translate. Does the order matter for multiplying the three different axis of rotation? I have three 4x4 matrices for x, y and z.
safe radishBOT
The order for transformation matrix multiplication is Scale, Rotate, Translate. ...
safe radishBOT
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Available help channel!
celest dawn
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flat frigateBOT
Civil Service Pigeon
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@celest dawn Has your question been resolved?
calm wagon
safe radishBOT
calm wagon
Does anyone know how to solve it?
heady lava
Helen
nocturne elm
Do you know how to find the area of a trapezoid?
nocturne elm
you don't need Heron here
nocturne elm
Actually maybe you do need Heron. It doesn't say that GC and CD are aligned
heady lava
XD
calm wagon
Do you know how to find the area of a trapezoid?
Nope
nocturne elm
You know how to find the area of a rectangle and a right-angled triangle, right?
nocturne elm
I do
Suppose we have this trapezoid. We can split it into two right-angled triangles and a rectangle. Then we can add up their areas to find the area of the whole trapezoid. Does that make sense?
nocturne elm
Yess
So the rectangle is: a * h, the left triangle is b * h / 2 and the right rectangle is c * h / 2. If you add them up:
ah + bh/2 + ch/2
h(a + b/2 + c/2)
h(a + (b + c)/2)
In other words, the area is: 1/2 * height * (top_side + bottom_side)
safe radishBOT
@calm wagon Has your question been resolved?
raw pivot
2pi periodic and continues
brave wolf
and
raw pivot
is that true?
brave wolf
That's true, but try f(x + pi)
thin bridge
technically yes, but there's a smaller period
brave wolf
Try graphing it
lean otter
think about how the modulus function affects other functions, when they're composed and imagine this on the graph of $\sin(x)$
flat frigateBOT
45
lean otter
so how does $g(x) = |x|$ affect some arbitrary $f(x)$ for $g(f(x))$
flat frigateBOT
45
brave wolf
hmm
lean otter
indeed
and what is the distance between the first point
and the endpoint of the first wave
differentiability is not the issue at hand
we're talking about periodicity
yes
correct
it's half the distance
of sin(x)
$f(x) = f(x + n\pi), n \in \mbb{Z}$
flat frigateBOT
45
lean otter
yes, since 2 is an integer
stark spruce
That will not help you in determining the period or continuity.
safe radishBOT
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Available help channel!
fickle grove
safe radishBOT
gentle edge
I don't really remember but don't you multiple probability every time you go to a new number
So won't it be .4x.4x.4x.4
fickle grove
let me try
fickle grove
uhhh
inner parrot
just change the 4 to a 3 for the 0.6 and it should be it
there’s only 7 trials
not 8
fickle grove
wait why 3 ?
inner parrot
so how many successes do we want
fickle grove
4
fickle grove
7
inner parrot
yw!
fickle grove
.close
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cloud shell
Could I get some assistance please
safe radishBOT
Could I get some assistance please
cloud shell
and how do I play chess with others?
spice wing
and how do I play chess with others?
Need to message them and make friends with them
spice wing
Could I get some assistance please
You see its a cubic graph , the graph will be below x axis when 2x+3>0 and else above y axis
You see its 0 at x = 1, -3/2
its the points where it takes turns because they r the roots
For more info i am providing the graph
,w graph y = -(2x+3)(x-1)²
flat frigateBOT
spice wing
you play chess yes?
yeah bro i used to play , had a rating of 1500
cloud shell
You see its a cubic graph , the graph will be below x axis when 2x+3>0 and else ...
I appreciate the assistance thank you
cloud shell
yeah bro i used to play , had a rating of 1500
wow that is pretty high do you still play for any chance?
You must be an intelligent individual
spice wing
wow that is pretty high do you still play for any chance?
Ive stopped playing
spice wing
wow that is pretty high do you still play for any chance?
Nah you will make it easily
Focus on playing with bots
And follow some book for strategies
You will rank up fast !
cloud shell
Ive stopped playing
I see I just got back into it after a decade of stopping my brother convinced me to I taught him how to play
cloud shell
You will rank up fast !
I will take the advice
spice wing
I see I just got back into it after a decade of stopping my brother convinced me...
Same for me , i stopped around 2 yr ago
safe radishBOT
@cloud shell Has your question been resolved?
upper jackal
binomial
safe radishBOT
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gusty raven
can anyone help
safe radishBOT
can anyone help
flat frigateBOT
jandro
split moss
Can you move forward from here?
safe radishBOT
@gusty raven Has your question been resolved?
safe radishBOT
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clever swift
royal kiln
Hmm..we can force the argument of the second function to be -6 again if x = -10
...and I think thats the whole problem, actually
safe radishBOT
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mossy ridge
How to solve
mossy ridge
btw which question?
The first qn rn
opal tartan
The first qn rn
think about the binomial expansion of (1+x)^n
(1+x)^n = nC0 + nC1 x^1 + ... + nCn x^n
integrate (x=0 to 1 to avoid constant)
(1+x)^n+1 / n+1 - 1/(n+1) = nC0 x + nC1 x^2/2 + ... + nCn x^(n+1)/(n+1)
put x = 2
I believe now you can solve it
mossy ridge
put x = 2
OHHHHH RIGHT
Makes sense
Thanks a lot
I dont like these kind of problems
opal tartan
I dont like these kind of problems
these are interesting problems if you try to think about the pattern in the series there