omg that was hard
#help-10
timid silo
high lily
(x+-2) can be simply written as (x-2)
timid silo
ooooh i see
high lily
and then you can apply zero product property (or null factor theorem or whatever its called where you're at) to get the solutions
polar harbor
oooh we only put it in brackets and thats it
Well you have to put brackets obviously for factorising
timid silo
Well you have to put brackets obviously for factorising
yea
timid silo
So what did you get?
yeh uhm well
it disappeared....
like i went back on it and then a different question came up
polar harbor
like i went back on it and then a different question came up
Huh?
polar harbor
like i went back on it and then a different question came up
Are you doing this online
timid silo
Are you doing this online
ahhh i relaised i have one online and paper copletly different the one i showed u guys was paper
sooo yeh its on paper but i think its right
polar harbor
sooo yeh its on paper but i think its right
You think mhmm
timid silo
You think mhmm
mhm hmm yupp shud be
polar harbor
mhm hmm yupp shud be
Ok then if you know how to find the solution and understand the steps
polar harbor
mhm hmm yupp shud be
Feel free to close the chat ☺️
timid silo
blush.
timid silo
Feel free to close the chat ☺️
how?....
timid silo
Ok then if you know how to find the solution and understand the steps
ahhh thank you
polar harbor
.close
agile plinth
$ln x = 3$
obtuse pebbleBOT
$ln x = 3$
warm shaleBOT
killua
agile plinth
we use ln for e so this would mean that x = e^3
or is this wrong?
vernal wren
It's correct
hexed vortex
that is corect
urban panther
quick question is it easier to calculate moles from volume or length?
obtuse pebbleBOT
quick question is it easier to calculate moles from volume or length?
long monolith
in chemistry?
obtuse pebbleBOT
@urban panther Has your question been resolved?
long monolith
volume then
long monolith
I have to choose one of these questions
,rotate
warm shaleBOT
long monolith
what's the question?
obtuse pebbleBOT
@urban panther Has your question been resolved?
little obsidian
i have a question about basic mathematics
obtuse pebbleBOT
i have a question about basic mathematics
little obsidian
and this is gonna seem really stupid but
lets say we have 6(1/4) = h/3
and we were trying to isolate h
we would multiply both sides of the equation by 3 right
to just get h on its own
my question is
for 6(1/4) when we multiply that by 3
why is it that we only multiply one of the two numbers by 3
i know like if it was 3r and we multiplied that by 3 we wouldnt multiply the 3 and the r by three separately we would multiply only one of the two by three to get 9r
i just want to know how exactly it works cos if it was 6 + 1/4 or 3 + r for example and we multiplied that by three we would multiply every number or term by 3
like whats the explanation behind it
little obsidian
can you expand on that?
obtuse pebbleBOT
@little obsidian Has your question been resolved?
Available help channel!
grizzled belfry
help
rotund flint
So I'm Graphing linear equations I believe, and we're like making standard form into y-intercept form or whatever it's called, but I know how to do the questions like (5x+2y=-4) perfectly fine.
But I'm always stuck on these
Types (-x+2y=8) or (-x + y=-4)
Those types, I really need to know the formula to these types so I can be able to do it each time if it's no problem, thanks for those that can help me out and explain
obtuse pebbleBOT
So I'm Graphing linear equations I believe, and we're like making standard form ...
timid silo
how do you do it
5x+2y=-4
cuz if u know how to do this
uhh there shouldnt be any problem
rotund flint
So it's exactly the same as the first one?
What do I do with the X and the Y?
Like how do I graph that?
timid silo
the same
rotund flint
Isolate the Y?
timid silo
ye
rotund flint
Would it be Y= 4 + x then
rotund flint
Wait so if both have negatives do you just not make them positive?
Like cause- ykw I'm overthinking it
rotund flint
Alrighty thx
timid silo
y=1/2 x -2
rotund flint
Wait what-
Wait lemme backtrack
Where the 1 come from
Oh wait, is that cause it can't be divided? To make it into a fraction like form or?
timid silo
y=1/2 x -2
this?
rotund flint
Yes
timid silo
1/2 = 1÷2
rotund flint
Yeah Ik, I'm just wondering where the 1 come from
timid silo
-3x+3y=3 for example
add 3x to both sides so u can isolate y
so u get 3y=3x+3
now divide 3
both sides
so u get the general form
rotund flint
I already know thissss- I just don't know that one I was talking about or how to solve it- math is so extra smh
kindred spoke
Anyone needs help?
rotund flint
Yes pls
Lemme repaste
So I'm Graphing linear equations I believe, and we're like making standard form into y-intercept form or whatever it's called, but I know how to do the questions like (5x+2y=-4) perfectly fine.
But I'm always stuck on these
Types (-x+2y=8) or (-x + y=-4)
Those types, I really need to know the formula to these types so I can be able to do it each time if it's no problem, thanks for those that can help me out and explain
This explains it all
kindred spoke
So I'm Graphing linear equations I believe, and we're like making standard form ...
hmm
rotund flint
?
kindred spoke
So I'm Graphing linear equations I believe, and we're like making standard form ...
What's wrong with this?
rotund flint
Idk how to do it is the problem -
kindred spoke
$y = \frac{8 + x}{2}$
warm shaleBOT
Science Done Right
rotund flint
No like my problem is ones like these, like I need the formula so I can be able to do some like them but one I have is
-x+y=-4
kindred spoke
No like my problem is ones like these, like I need the formula so I can be able ...
To solve the equation for x and y?
rotund flint
Wait one sec
kindred spoke
What exactly is your goal
rotund flint
The goal for school or for this equation?
Is this confusing or something? I'm just asking for the formula to equations such as those
kindred spoke
Is this confusing or something? I'm just asking for the formula to equations suc...
What formula??
rotund flint
rotund flint
What- but I said exactly what it says
Ik how to solve the top problem
But idk how to do ones like the one at the bottom
kindred spoke
Should I solve and show the solution?
rotund flint
Yes pls
rotund flint
Lol
kindred spoke
the equations are simple, as in the coefficients are very small, so try eliminating one variable
it's completely valid btw
rotund flint
I just needed to know what -x + y = -4 is, I'm Graphing
Like how to do it-
Nvm smh thanks tho lolll
obtuse pebbleBOT
@rotund flint Has your question been resolved?
fierce lagoon
Okay so @gloomy cove
obtuse pebbleBOT
Okay so <@680964817737351236>
fierce lagoon
Lemme show you what they mean by that
Let's say we have $\int_{e}^{e^2}\frac{\ln{(x)}}{x}{dx}$
warm shaleBOT
Umbraleviathan
warm shaleBOT
Umbraleviathan
fierce lagoon
So that $du = \frac{1}{x}{dx}$
warm shaleBOT
Umbraleviathan
fierce lagoon
The problem is that when we substitute, we are evaluating it with the wrong bounds
We need to change them from x-bounds to u-bounds
warm shaleBOT
Umbraleviathan
gloomy cove
X = 1
fierce lagoon
So you plug the x-bounds into what u is
fierce lagoon
**Umbraleviathan**
Which is here
ln(e) = 1
ln(e^2) = 2
So the integral becomes
$\int_{1}^{2}udu$
warm shaleBOT
Umbraleviathan
fierce lagoon
**Umbraleviathan**
From this
gloomy cove
Ohh ok
fierce lagoon
**Umbraleviathan**
To here
We have to put the x-bounds into "u" to get the u bounds
You understand?
gloomy cove
Ok
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
crystal stream
Ok
crystal stream
kind of a silly question
regarding inequalities
Ok
So I am trying to find a domain of convergence for this one and copute its value as a formula of x
So if |t|<1
I wanted to know if my steps are correct from there
And actually from the last line I don't know how to continue
<@&286206848099549185>
obtuse pebbleBOT
@crystal stream Has your question been resolved?
crystal stream
<@&286206848099549185>
obtuse pebbleBOT
@crystal stream Has your question been resolved?
obtuse pebbleBOT
@crystal stream Has your question been resolved?
open sorrel
the inequalities below can be condensed
$|1-x|^2>1 \implies |1-x|>1 \implies 1 - x > 1 \text{ or } 1 - x < -1 \dots$
warm shaleBOT
vin100
obtuse pebbleBOT
@crystal stream Has your question been resolved?
tame valve
Is the constant in an exponential equation equal to the asymptote?
obtuse pebbleBOT
Is the constant in an exponential equation equal to the asymptote?
obsidian isle
can e^x ever be 0?
tame valve
can e^x ever be 0?
sry i didnt see
no it cant
my discord crashed
i was gonna say some other stuff
tame valve
Is the constant in an exponential equation equal to the asymptote?
I know the asymptote in a logarithmic equation is x = c where $y = log_b(a(x - c)) + d$
I'm watching the organic chemistry tutor and he seems to imply that although I'm not 100% sure.
warm shaleBOT
Jowo Rinpoche (FSM!!)
tame valve
Is the constant in an exponential equation equal to the asymptote?
My question if the constant was always the asymptote. I can see that e^x can only get progressively closer to 0 and not reach it
But what about other scenarios like multiplication/etc?
does this apply in every scenario
obsidian isle
My question if the constant was always the asymptote. I can see that e^x can onl...
If e^x can't approach 0 then e^x - 1 can't approach -1
So yes
The constant term is the asymptote
tame valve
If e^x can't approach 0 then e^x - 1 can't approach -1
ohh ok
got it, thx!
@obtuse pebble
.close
twilit wagon
how can i solve by sustitution
obtuse pebbleBOT
how can i solve by sustitution
winter ether
how can i solve by sustitution
theres no composite function so we wouldn't use usub, instead integration by parts I believe
obtuse pebbleBOT
@twilit wagon Has your question been resolved?
winter ether
@twilit wagon what do you still need help with? Do you know integration by parts?
obtuse pebbleBOT
@twilit wagon Has your question been resolved?
timid silo
obtuse pebbleBOT
brave bramble
Is that f times g?
brave bramble
Then I suggest looking at
f(-x)g(-x)
timid silo
Then I suggest looking at
f(-x)g(-x)
ok
brave bramble
And simplifying it, given f is even and g is odd
timid silo
And simplifying it, given f is even and g is odd
are
(f*g)(x) and f(g(x)) the same right
brave bramble
not f of g
This implies to me that, no, this does not mean f(g(x))
But I don't know for sure, haha.
f(g(x)) is normally given by f°g
timid silo
yea
so
im confused for what do to
to do*
because if if is even
lets say 2
2 * 1 = 2
or
4*1 = 4
brave bramble
That's an even number, not an even function
brave bramble
Then I suggest looking at
f(-x)g(-x)
And simplifying it, given f is even and g is odd
brave bramble
It simplifies as such:
(f•g)(-x)
= f(-x)g(-x)
= f(x)(-g(x))
= -(f•g)(x)
timid silo
It simplifies as such:
(f•g)(-x)
= f(-x)g(-x)
= f(x)(-g(x))
= -(f•g)(x)
im still confused
brave bramble
No problem what's up
timid silo
No problem what's up
im confused how i
simplify theproblem
could i plug any number into
x
brave bramble
Maybe you have an alternative solution than I do. Show it off!
I'll let you know if it works
brave bramble
Okay, let me know what's weird
timid silo
Okay, let me know what's weird
so lets say
(f*g)(-2)
= f(-2)g(-2)
=f(2)(-g(2))
=-(f*g)(2)
so
if i put 2 in for x
like this righ
then it becomes even?
brave bramble
What becomes even?
timid silo
What becomes even?
the function right
brave bramble
f*g is odd
timid silo
f*g is odd
im confused how
brave bramble
Because
(f•g)(-x)
= f(-x)g(-x)
= f(x)(-g(x))
= -(f•g)(x)
timid silo
lok
brave bramble
What do we need to prove, in order to show f*g is odd?
brave bramble
Why g?
brave bramble
You're guessing
We need to show that
(f•g)(-x) = -f•g(x)
It might help to just call f•g by a different name. Let's call it h!
timid silo
oh
brave bramble
Then in order to show that h is odd, we want to show
h(-x) = -h(x)
mossy hull
hi
mossy hull
my work is
ratio test
thats the first time
this is the second term
crossing out all the 2n!
and rewriting the base 3 expressions as 1/9
mossy hull
and the limit of this expression is going to be 1/9 because ratio of leading terms so L=1/9<1 so the series is convergent yes?
obtuse pebbleBOT
@mossy hull Has your question been resolved?
dreamy scroll
between these two steps
mossy hull
hmm?
dreamy scroll
you have too many terms on top? $$\frac{(2n+4)!(2n)!}{(2n+2)!(2n+2)!} = \frac{(2n+4)(2n+3)}{(2n+2)(2n+1)}$$
warm shaleBOT
JamesH
dreamy scroll
oh wait
mossy hull
yeah I did kinda overexpeand the factorials
dreamy scroll
i may have misread
mossy hull
OH yeah let me explain
dreamy scroll
i think what you wrote makes sense but just wasnt efficient
mossy hull
yeahhhhh
I know
I just wanted to make sure I was crossing out terms for the right reasons
so I was like "lets make it all go to 2n!"
that way I couldn't make a mistake
dreamy scroll
fair enough
i agree with your reasoning, the limit will be 1/9 so the series will converge
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
obsidian tusk
yo
obsidian tusk
<@&286206848099549185>
civic zealot
read this.
nocturne minnow
<@&286206848099549185>
Two things, one you have to wait at least 15 minutes before pinging helpers and two you never posted a question you need help on
urban patrol
use the formula for a sphere
obsidian tusk
idk
nocturne minnow
Hmmmm... It's a quiz, are you allowed to get outside help on this?
obsidian tusk
Hmmmm... It's a quiz, are you allowed to get outside help on this?
no
its not a quiz
its just iready
nocturne minnow
Then why does it say quiz in the title?
nocturne minnow
Well, apply the formula for volume of a sphere
obsidian tusk
idk how to
short yarrow
idk how to
substitute the value
obsidian tusk
dk how to?
short yarrow
hmm
nocturne minnow
idk how to
Look up the formula
obsidian tusk
bro
short yarrow
just apply the val of radius
obsidian tusk
i cant even go to school
short yarrow
sed
nocturne minnow
As I mentioned, if you don't know what the formula for volume of a sphere is, look it up
Then plug in the values into the formula
obtuse pebbleBOT
@obsidian tusk Has your question been resolved?
placid frigate
obtuse pebbleBOT
placid frigate
twin sapphire
so how do you understand limits and continuity?
placid frigate
something like where they meet?
placid frigate
i think this is 4
placid frigate
idk about this one
twin sapphire
i think this is 4
how can 4 be the answer to "does something exist?"
obtuse pebbleBOT
@placid frigate Has your question been resolved?
hollow sparrow
obtuse pebbleBOT
hollow sparrow
I'm trying to find the areas of A and B,
I tried changing x=(y-1)^2-2 to a curve based on x
but it will give me ∓
so I'm not sure how to integrate to find the area
twin sapphire
try to find the points where the parabolas cross
hollow sparrow
twin sapphire
i meant a formula for them
anyway after you have them
these will be the boundary points for your integrals
to integrate a
express x in function of y
for both curves
this way you can do an integrate with y as the parameter
hollow sparrow
I think I'm getting it, so I will integrate twice, and use dx for one and dy for the other
is that correct?
twin sapphire
yeah but also be wary
when you will express y in function of x
you'll get two possibilities
since the function is not inversible
pick the one that corresponds to the domain you're interested in
like for example for one y there are 2 possible x for the blue curve
but you are interested in only the negative xs for A
hollow sparrow
I see
twin sapphire
dont you know a formula for tan(a+b)?
which you could use to express tan(2theta) as tan(theta + theta)
haughty sail
theta + theta?
hollow sparrow
where can I calculate a definite integral in dy
twin sapphire
all the integrals are definite here i think
hollow sparrow
if I integrate the same format but swap the y for x will it give me the same value?
twin sapphire
oh if you meant this
yeah those two are the exact same thing
you just changed symbols
hollow sparrow
but the axis are different no?
twin sapphire
well,
the two integrals are the same
like the result of this integral is linked to the geometry and the curves
but the integral itself doesnt care
how you write the variable inside it
shell pine
obtuse pebbleBOT
shell pine
ik that 49 squared - 33 squared is 511
and that 511 sqaure rooted is 22.60
but how do i find the other sides base
hollow sparrow
wait so do both the triangles have the same base
hollow sparrow
nice
shell pine
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
static sun
obtuse pebbleBOT
tranquil arch
expand LHS
tranquil arch
and tan(pi/4)=1
static sun
Oh god
I can't believe I forgot that
Hahaha
And then make a common denominator?
tranquil arch
yes
tranquil arch
$\frac{1+t}{1-t}-\frac{1-t}{1+t} = \frac{(1+t)^2}{(1-t)(1+t)} - \frac{(1-t)^2}{(1+t)(1-t)}$
warm shaleBOT
秋水
static sun
Yeaa
It ended up giving me 2tana/1-tana^2
So we kewl
Yea I gotta do that
Thank you
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
obtuse pebbleBOT
timid silo
how do I solve this
I initially thought the line with the angle 72 was traversal but when checking the answer key my answers differed a lot
where would I start>
silk shadow
interior angles: a = 72
silk shadow
🧐
worn ocean
It's 72
The answer key is wrong
because the lines are parallel, opposite angles are always equal on an intersecting line
obtuse pebbleBOT
@timid silo Has your question been resolved?
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
a
timid silo
answer key says 60
kindred spoke
Check the question again, are you sure you have to find a?
kindred spoke
Oh I see
timid silo
I got 72 also but answer key says it's 60 degrees
kindred spoke
wtf?
timid silo
yeah
timid silo
😂 that's the key they gave us so idk
kindred spoke
Is this an exam? If yes request your teacher to change the key
timid silo
no this is just homework
timid silo
ig
kindred spoke
I will go to help another person
timid silo
if the sum of 11 consecutive even numbers is 'm' then the largest of the numbers in sum of 'm' is
obtuse pebbleBOT
if the sum of 11 consecutive even numbers is 'm' then the largest of the numbers...
timid silo
$a) \frac{m}{11} + 11 \\
b) \frac{m}{11} + 12 \\
c) \frac{m}{11} + 10 \\
d) \frac{m}{11} + 9$
warm shaleBOT
kube
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
<@&286206848099549185>
kindred spoke
if the sum of 11 consecutive even numbers is 'm' then the largest of the numbers...
I will try to send a solution
timid silo
ok
kindred spoke
because we have only one equation with m
so try making it look similar to that previous equation containing m
kindred spoke
this will help you make it in terms of m, which is what is given in the options
heavy wedge
obtuse pebbleBOT
heavy wedge
hi
im stuck on part c
i got a and b right
the answer for a and b is 64.34m and 53.98m respectively
<@&286206848099549185>
lusty cedar
see the triangle with difference of heights and distance between the buildings as the triangle
use arctan
mighty ether
Hi guys I have a question about infinte series
obtuse pebbleBOT
Hi guys I have a question about infinte series
mighty ether
if you can find the formula for the sum of an infinite series
can you just take the limit of the sequence at infinity, to get the value the series converges to?
eg:
*value the series convergs to
copper marten
you mean you have found a formula for the sum of a finite series and then want to expand it to an infinite series by taking its limit to inf?
copper marten
yes
mighty ether
is it correct to call this the formula for the partial sum?
and can I take the limit of this at n = infinity to obtain the value it converges to?
copper marten
if your sum is $\sum_{k=2}^{n} \frac{(k)!-1}{(k)!}$ then yes
warm shaleBOT
Rœmer
copper marten
i think your misunderstanding something
mighty ether
probably
copper marten
**Rœmer**
if you take the limit $k$ to $n$ of this, the sum doesnt converge to that, but that is just the n-th term of the sum
warm shaleBOT
Rœmer
copper marten
you first need to find a formula for the value the sum converges to
mighty ether
this was the series originally
so we did actually calculate the first sums
and then this formula for the nth term
mighty ether
but don't we have to sum them ?
so i got n = 1, n= 2, n =3 and n=4 but then
we have to actually sum them don't we?
copper marten
no but you're summing the sums (if that makes sense)
mighty ether
i'm soo confused because I originally assumed the sums as well, like you said
but pretty much everyone else has actually summed them
copper marten
but pretty much everyone else has actually summed them
that's incorrect
warm shaleBOT
Rœmer
copper marten
so you shouldnt sum 1/2 and 5/6 again
copper marten
then this formula indeed holds
so $\sum_{k = 1}^{n} \frac{k}{(k+1)!} = \frac{(n+1)!-1}{(n+1)!} = 1- \frac{1}{(n+1)!}$
wait no
and now you can take the limit of n to infinity
warm shaleBOT
Rœmer
copper marten
and i think you can see what this converges to
mighty ether
yes i can i'm just still confsued about the last few steps, i think maybe this is a point of terminology here?
liek we wouldn't call taht the formula for the partial sum?
idk
copper marten
yes, the $\frac{(n+1)!-1}{(n+1)!}$ is the formula for the partial sum
warm shaleBOT
Rœmer
mighty ether
okay and once we get a formula for the partial sums
we can always take the limit at infinity to see what our infinite series converges to?
copper marten
yes (or diverges)
copper marten
great
mighty ether
im just a little unsure why we weren't meant to actually "sum" before
you said it was incorrect
copper marten
as in the last step here?
warm shaleBOT
Rœmer
mighty ether
so im thining maybe we should have a comma instead of a plus?
to show we arent actually summing
but just listing the values of our series?
warm shaleBOT
Rœmer
copper marten
$p(2) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
warm shaleBOT
Rœmer
copper marten
etc.
copper marten
yes
mighty ether
makes sense
thank you so much
so basically, if we get something in summatino notation and its our infintie sum
if we can figure out some pattern that gives us the partial sums
then we can take the limit at infinity to see whether the sequence diverges/convergres
but normally it is obviously quite tedious
copper marten
if we can figure out some pattern that gives us the partial sums
well, just the value of the sum for all n
mighty ether
mhm
copper marten
but normally it is obviously quite tedious
yes, a lot of times that isn't possible
copper marten
np!
mighty ether
also, out of curiousity
would we have been able to get the sum using the method for telescoping sums?
i just saw the minus sign there and thought of telescoping sums but
could be way off the mark there
.close
worthy glacier
how do i do this
obtuse pebbleBOT
how do i do this
worthy glacier
just apply pythagoras
how do i do that
pine sail
how do i do that
Do you know what the theorem is?
worthy glacier
Do you know what the theorem is?
a2 +b2 =c2
worthy glacier
yeah
pine sail
Find c.
worthy glacier
how do i know if base is correct
so its
122 + 122?
12 squared plus 12 squared
pine sail
How did you get 12^2 = 122?
rough flare
How did you get 12^2 = 122?
he doesnt know how to do ^2 sign
so he meant 12 squared plus 12 squared
pine sail
That's not the point.
rough flare
12 squared plus 12 squared
but yh u root that for c
pine sail
Where did 122 come from?
rough flare
Where did 122 come from?
he means 12^2
but he didnt do the ^2 sign cause he doesnt know how to
pine sail
I'm not talking to you.
For god's sake before barging in, look what I mean.
12^2 or 12 squared if that's what you prefer.
Is not 122
rough flare
bro
pine sail
,calc 12*12
warm shaleBOT
Result:
144
r u even reading what he said he cant do it
he meant 12 squared tho
thats why he typed it afterward
robust sleet
pine sail
<:sippy:920835242980442113>
You aren't a helper anymore?
robust sleet
I still help but role was useless coz I don't look at tags
warm canopy
I think the point miniben was making was that they were trying to write 12^2, but missed the ^ symbol
rough flare
I think the point miniben was making was that they were trying to write 12^2, bu...
yh i was saying that but bro is so tunnelvisioned or smthn
pine sail
And my point was even if they write 12 squared that doesn't change the fact it's not 122.
worthy glacier
but he didnt do the ^2 sign cause he doesnt know how to
yep
warm canopy
And my point was even if they write 12 squared that doesn't change the fact it's...
OP never claimed 12^2 was 122
pine sail
You write 12^2
worthy glacier
o ok
pine sail
For 12 squared.
worthy glacier
sorry for the confusion didnt knoow
pine sail
Nevermind, I suppose it was my bad.
worthy glacier
it ok
robust sleet
Even though they didn't depict the triangle well, if they meant that the base and the height are same length then there should be that small stripe at the height too, not only at the base, which confuses me.
pine sail
Even though they didn't depict the triangle well, if they meant that the base an...
That's right.
It just shows its equal to something. But doesn't tell anything about that something.
So presumably it's the height.
robust sleet
Otherwise with one side known its impossible to find the unknown side, so most likely base is 12 and they forgot to stripe the height.
worthy glacier
so top and bottom is c
robust sleet
for the exercise at right yes
worthy glacier
12 is on left
robust sleet
a and b can either be 9 or 12 doesn't matter
worthy glacier
a^2 + b^2 =c^2?
robust sleet
yes
worthy glacier
thats it?
robust sleet
yes
worthy glacier
ok
obtuse pebbleBOT
@worthy glacier Has your question been resolved?
valid crown
Is there a relatively simple way of generating all possible binary combinations without ever repeating ones?
obtuse pebbleBOT
Is there a relatively simple way of generating all possible binary combinations ...
valid crown
I code things in matlab and sought way for optimizing things but computer is burning for big number of digits, and as it appears, most of them aren't even usefull
checked for 8 digits, and combinations without repeating come up to 51 possibilites
which is 10% of all possibilites
I found a way to make all combinations BY HAND you just use either:
0 at the beginning
1 at the end
all options of 1 and 0 like: 10; 100; 1000; and then move them accordingly
maximum number of ones is half a digits
timid silo
what
valid crown
here is how it looks:
all combinations of 1 and 0 that never duplicate 1
I know I can code a chunk, then convert by digits
kind hawk
hmm my best idea is to do it recursively. the string will start with some number of 0, then a 1, then a 0 and then something
valid crown
but then it won't be efficient
kind hawk
this something you can calculate recursively how many options there are
so for example if you have 10 digits, then a part of that starts with 10 and then all valid combinations of 8 digits
then 010 and all combinations with 7 digits
etc
valid crown
yeah, I know this way but...
kind hawk
if you don't actually program it recursively I think this should be somewhat efficient
valid crown
if I reconvert it to a matrix 010 -> 0; 1; 0; it could take lot of time
I wondered if there was a more... clever way
timid silo
like i dont know much about matlab and its possibilities, but if you really want it be really efficient and not make any unneeded calculations then, just do a for loop and put 1 in every position, then do double nested loop where it goes from i=0 to 7 and other from j=i+1 to 8, then do 3 nested... it will only get as many possible combinations as there is
valid crown
oh and on the image: I divided parts to chunks
kind hawk
not sure what you mean by reconverting it to a matrix?
valid crown
you said to check by a string right?
kind hawk
well string, vector, same thing
valid crown
but then I need to paste all sets of valid combinations to 2-dim array
and for all of those chunks, 0;1;10;100... and so on there will be eparate loop if it's big enough
then another one for all permutations
I wondered if there is a way to approach this from n choose k
kind hawk
permutations? not sure what you mean. you can't just permute a combination or you might end up getting a repeated 1
also not sure what yo mean with chunk
valid crown
you can
IF
you apply a rule
0 at the beginning
1 at the end
and between
10, 100, 1000, 10000 and so on
to figure whether those that I listed are ALL possibilites I made those forementioned chunks
see 2nd column first row
1 chunk consisting 3 combinations
10101000
10100010
10001010
kind hawk
ah that is what you mean
valid crown
It's made of only parts that I listed
2x 10 and 1000
off course 8 digits is a small number
so besides 0,1 at both ends
there are usually 3 parts
kind hawk
this seems like it boils down to roughly "how many partitions of a number are there" which is not an easy question
valid crown
so no much permutations to shuffle them around
but with big number, it will take time
Yeah, but it was mine way of seeing problem
thought there was another approach
in the end I don't need to know how many chunks
just 1x 2-dim array of all possible combinations depending on the number of digits
If it helps it may be non repeating zeros in a combination
Might as well just negate it at the end
kind hawk
hmm yeah dunno if there is a good way to do it. I would probably do it recursively in the way I described
valid crown
I think there might be a solution in a pascal traingle, to analyticaly contract the code
but I'm not as fast in figuring all the ways it works
kind hawk
let's ask a different question, why do you want to do this? just an assignment? fun?
valid crown
for something even harder to comprehend
it's not even the hard part yet
but I thought I'll do most of the work myself
kind hawk
well maybe you are tackling that other problem wrong
kind hawk
and this is an example of a XY problem where you ask for a solution to Y even though you actually want to ask for a solution to X
valid crown
more like X(Y)
if so, yeah
but the X problem may not be yet solved
so I'm digging
I meant, there may be nowhere a proof that it is solvable
timid silo
will you only do it for n=8?
valid crown
but no 1 repetition in a combination seems kinda easy
yeah
depending on digits number
but I just found out that for n=8 it decreases all VALID possibilites to a one of tenth
if I run a code on n=20 it calculates for hours with no answer
using nearly all of RAM
timid silo
then you wrote really shitty code if it cant check 2^20 possible options in few seconds
valid crown
no, it's just hard to predict which of those would be Valuable
seemingly non important choice early may have big impact later on
- all of this is just for me the check if there are more patterns
kind hawk
the point is that even the most stupid way of calculating every combination and removing all invalid ones shouldn't take hours
valid crown
it only changes values of 2 variables
but for n digit
it does n*(2^n) combinations
timid silo
can you show what you wrote
valid crown
sure
clear all;
%hold on;
% initial condition
steps = 10;
combinations = 2^steps;
gain = zeros(steps+1,combinations);
gain(1,:) = 1;
growth = zeros(steps+1,combinations);
choice = dec2bin(0:2^steps-1)' - '0';
%condition embedded in mathematical operations+
for i = 1:combinations
for j = 1:steps
gain(j+1,i) = 0;
growth(j+1,i) = growth(j,i) + choice(j,i)gain(j,i);
gain(j+1,i) = not(choice(j,i))(gain(j,i) + growth(j,i));
end
end
subplot(2,1,1);
plot(gain);
subplot(2,1,2);
plot(growth);
%condition check
% for i = 1:2^(steps)
% for j = 1:steps
% if(choice(j,i))
% growth(j+1,i) = growth(j,i) + gain(j,i);
% gain(j+1,i) = 0;
% else
% growth(j+1,i) = growth(j,i);
% gain(j+1,i) = gain(j,i) + growth(j,i);
% end
% end
% end
I'm trying to find way of maximizing gain or growth in a giving choice frame interval
if there is no choice DELAY for buffing growth with gain, then the solution is too TRIVIAL
but otherwise it is pointless to buff growth 2x times in a row before it produces any gain
this thing here is still pretty... "easy"
for I wanted to include some coefficients of value assignment
and further transform into continous analogue
checking it for simple functions like log,exp,poly etc to generate assignment coefficients
subplot(2,1,1);
plot(gain(:,100));
subplot(2,1,2);
plot(growth(:,100));
you can also use code up here to see that some combinations are just... garbage
kind hawk
hmm, are you sure there are 51 combinations for n=8 ?
I have 55 and I don't think I double counted something
valid crown
could you pass it please?
timid silo
there cant be 0 1's?
kind hawk
10101010
10101001
10100100
10100101
10100010
10100001
10100000
10010100
10010101
10010010
10010001
10010000
10001000
10001010
10001001
10000100
10000101
10000010
10000001
10000000
01010100
01010101
01010010
01010001
01010000
01001000
01001010
01001001
01000100
01000101
01000010
01000001
01000000
00101000
00101010
00101001
00100100
00100101
00100010
00100001
00100000
00010100
00010101
00010010
00010001
00010000
00001000
00001010
00001001
00000100
00000101
00000010
00000001
00000000```hmmm
timid silo
yeah i am also getting 55
kind hawk
used code:
A = cell(n, 1);
A{1} = [0;
1];
A{2} = [0, 0;
1, 0;
0, 1];
for j = 3:n
s = size(A{j-2}, 1);
B = [ones(s, 1), zeros(s, 1), A{j-2}];
for k = 2:j-2
s = size(A{j-k-1}, 1);
B = [B;
zeros(s, k-1), ones(s, 1), zeros(s, 1), A{j-k-1}];
end
B = [B;
zeros(1, j-2), 1, 0;
zeros(1, j-2), 0, 1;
zeros(1, j-2), 0, 0];
A{j} = B;
end```runs fast enough at least for small inputs
most of the time is probably spent copying the matrices so that's not very efficient
valid crown
ok I found out
chunk was migging
10101001
10100101
10010101
and trivial 00000000
hence 4 combinations more
thanks
kind hawk
if you just want to count them that's way faster
valid crown
welp your code would still be better than what I could came up with
In my case there'd be triple loop
you jut use it twice
I'll just check if the combinations are valid
timid silo
used code:
```n = 20;
A = cell(n, 1);
A{1} = [0;
1];
A{2} ...
idk much about matlabs but is it possible to convert this cpp code to matlab code? would it run faster? or is too hard?
kind hawk
I have no clue how the speed of programming languages really compares. I don't think matlab is super fast for stuff like this
I don't really know what your code does. but one thing you might want to change is the recursive nature
here you always call the same function again at some point during the iteration
timid silo
it generates all possible combinations for him
kind hawk
call it once and then save the result
timid silo
call it once and then save the result
? why would i
kind hawk
so unless I'm missing something, for many calls of generate(s, i, count) you are calling generate(s, i+1, count) twice
and so generate(s, i+2, count) four times
and generate(s, i+3 count) eight times
and so on
at low number of iterations this isn't that important but later on it might
for example try to implement fib(n) = fib(n-1) + f(n-2) either recursively like that or with a look up table which saves all the values
the recursive version sucks. for n=20 you are calling f(19) and f(18). for f(19) you are calling f(18) and f(17), so that's already two times calling f(18). and then so on and so on and at the end you are calling millions of functions
kind hawk
whatever that graph is