#help-0
ocean sealBOT
minor needle
proper quiver
slender prairie
lone heartBOT
quasi vector
,rcw
ocean sealBOT
quasi vector
how did you do the third step
slender prairie
I’m not sure if it’s correct but I separate it the absolute value
Like the 2 possible answers
Does that make sense?
quasi vector
you kept the same bounds of integration
plus the 2 will be distributed to both integrals
slender prairie
I’m not sure how to change it
slender prairie
plus the 2 will be distributed to both integrals
Yea I noticed that now that I’m looking at it😭
quasi vector
I’m not sure how to change it
for what values of x will x-2 be negative?
slender prairie
Anything less than 2?
quasi vector
upto 2
quasi vector
yes
lone heartBOT
@slender prairie Has your question been resolved?
safe mountain
this fine ?
lone heartBOT
this fine ?
alpine sable
hi
safe mountain
looks odd
alpine sable
@limber cradle
safe mountain
its my channel
quasi vector
this fine ?
where did the other dy/dx go?
quasi vector
i think you will have to substitute the inner function for a secondary variable
i.e u = x - y
for example
safe mountain
where did the other dy/dx go?
i tried to take it out i think, but i didnt take into account the 4
safe mountain
i.e u = x - y
thats what i did
jagged cobalt
how did you get to 4dy/dx(x-y)^3 - dy/dx=-3
should it not be 4dy/dx(x-y)^3+dy/dx=3
safe mountain
safe mountain
how did you get to 4dy/dx(x-y)^3 - dy/dx=-3
should it not be 4dy/dx(x-y)^3+dy/dx...
i moved over the dy/dx from the left hand side to the right
making it negative
jagged cobalt
i moved over the dy/dx from the left hand side to the right
you mean you moved the dy/dx on the right to the LHS via subtraction? because even then your answer isnt right - look at it
safe mountain
i just realised i fcked up from the start
jagged cobalt
subtract 1 from each side and you get 3-4 dy/dx(x-y)^3=dy/dx
move the dy/dx to one side and you have 3=4 dy/dx(x-y)^3+dy/dx
jagged cobalt
expand the left hand side (expand the 1-dy/dx)
move the dy/dx terms to one side and factor
safe mountain
expand the left hand side (expand the 1-dy/dx)
move the dy/dx terms to one side ...
yep
but i still have to divide
jagged cobalt
thats not what i meant
4(x-y)^3 (1-dy/dx)=4(x-y)^3-4dy/dx (x-y)^3
btw, what is u supposed to be here exactly
safe mountain
btw, what is u supposed to be here exactly
this the question
safe mountain
if youre doing the chain rule then u=x-y
yea thats what i did then i got du/dx = 1 - dy/dx
safe mountain
4(x-y)^3 (1-dy/dx)=4(x-y)^3-4dy/dx (x-y)^3
sry im confused as to what u did here
jagged cobalt
ahah, okay, its just here you have written u in the exponent?
also if g(u)=u^3 for your chain rule then dg/du=3u^2 not 4u^3
jagged cobalt
sry im confused as to what u did here
i just expanded the (1-dy/dx) bracket in the same sense that A(1-x)=A-Ax
safe mountain
ahah, okay, its just here you have written u in the exponent?
i only use u in the first part when doing chain rule and then i just use x & y throughtout
jagged cobalt
also if g(u)=u^3 for your chain rule then dg/du=3u^2 not 4u^3
yes i know, but theres no reason for u to be in the exponent unless you just wrote it by accident instead of 4
and this ^ still applies - you did the chain rule wrong
ah wait
nvm i see, my bad
jagged cobalt
also if g(u)=u^3 for your chain rule then dg/du=3u^2 not 4u^3
ignore this ^, the u in the exponent thing is still weird though
jagged cobalt
i just expanded the (1-dy/dx) bracket in the same sense that A(1-x)=A-Ax
you just need to do this so you can isolate the dy/dx terms
safe mountain
4(x-y)^3 (1-dy/dx)=4(x-y)^3-4dy/dx (x-y)^3
okk this kinda makes sense, imma try work with it
think i got
i dont get why that expansion is so hard to recognise tho
prolly rusty
lone heartBOT
@safe mountain Has your question been resolved?
warped sentinel
lone heartBOT
warped sentinel
this
tribal palm
lone heartBOT
tribal palm
Can somebody walk me through this
prisma badger
ok
mystic orchid
Do you know the trig ratios
prisma badger
must I do it or you
mystic orchid
Do you know the trig ratios
?
tribal palm
yeah
wary stream
Do you know what SOH CAH TOA is?
tribal palm
Yes
wary stream
At angle A, what are the opposite, adjacent, and hypotenuse sides?
tribal palm
C is hyp, b is opp
prisma badger
wary stream
C is hyp, b is opp
And what about opposite?
tribal palm
A is adj
wary stream
No
tribal palm
Oh wait I’m dumb I was right before
wary stream
You are looking for the ratios based on what angle it asks for
It's asking about angle A
tribal palm
12/20
prisma badger
sub the figures to get your answer
prisma badger
cos
tribal palm
💀
wary stream
No
You are looking for the ratios based on what angle it asks for
You are using angle A
Where is angle A in the diagram?
tribal palm
Sin(A)= 3/5
tribal palm
tan a is 0.01 I think
strange meadow
draw a triangle
wary stream
What?
wary stream
@strange meadow
strange meadow
Ignore what I said sorry
tribal palm
Sine cosine and tangent
strange meadow
Ok first do you understand why sinA =12/20
wary stream
tan a is 0.01 I think
Let's reset, you have the triangle and you are looking for sine, cosine, and tangent of A for that triangle. Based on where angle A is, what is the value for opp, adj, and hyp?
tribal palm
cause 12 is opp and 20 is hyp
tribal palm
16
tribal palm
Opposite over adjacent
strange meadow
Yes but the numbers
tribal palm
12 over 16
tribal palm
3/4
tribal palm
16/20
wary stream
Good, now simplify that
tribal palm
4/5
wary stream
Wait
Hang on
There's a slight difference with that
It's asking for cos B
Not cos A
wary stream
Yeah cosA was correct, but that wasn't a problem that it asked for
wary stream
Sin(A)= 3/5
.
tribal palm
3/5 again
wary stream
And that's it
tribal palm
Oh alright that’s easier than I was thinking
wary stream
You just needed to apply SOH CAH TOA, in perspective of the angle it wanted
fringe coyote
Hey I have a question about a question
lone heartBOT
Hey I have a question about a question
fringe coyote
Numbers from 0 to 50 are written on the cards. The cards are turned over, shuffled and one card is drawn without looking.
- What is the probability that the number written on the drawn card is two-digit?
And the correct answer is 41/51
I don't get it why is it 51 and not 50?
molten pivot
Ye
molten pivot
Listen
lone heartBOT
fringe coyote
.reopen
lone heartBOT
✅
molten pivot
Go look up what an "off by one" error is
molten pivot
This is actually a really common problem in a lot of domains and its important to understand the issue conceptually
fringe coyote
.close
errant umbra
someone help pls
lone heartBOT
someone help pls
alpine sable
similar triangles
weary abyss
can someone help me with the expression
lone heartBOT
can someone help me with the expression
weary abyss
> using the definition of derivative
i got the derivative, plugged in 5, and made it equal to f(x) but still wrong answer
i also tried without plugging it in got -5/6 and still wrong
alpine sable
limit definition of the derivative
weary abyss
what do i do with the limit definition
alpine sable
write it out in that box...?
weary abyss
(f(x+h) - f(x))/h ?
weary abyss
substitute x or h
alpine sable
h
weary abyss
so sub h with 5?
ocean sealBOT
Toby
alpine sable
we want $\lim_{x\to 5}$ of something
ocean sealBOT
Toby
alpine sable
note that as h goes to 0, x+h goes to 5
weary abyss
so what do i do
alpine sable
almost
weary abyss
wdym
alpine sable
now substitute x=5+h and rewrite the limit
weary abyss
???
alpine sable
how did you do the first part of the question?
its continuing on from that part
since we want a limit x-> 5 that evaluates to the limit
we have
$\lim_{h\to 0} \frac{f(5+h)-f(5)}{h}=\lim_{x\to 5}\frac{f(x)-f(5)}{x-5}$
ocean sealBOT
Toby
alpine sable
because we want f'(5)
weary abyss
so (sqrt(4+3x) - sqrt19)/(sqrt(4+3x) - 5)
weary abyss
okay thank you
alpine sable
np
frail jasper
How do i prove that the function is O(n^2)?
lone heartBOT
How do i prove that the function is O(n^2)?
alpine sable
think about the behaviour of ceil(n)
regal wadi
lone heartBOT
@frail jasper Has your question been resolved?
lone heartBOT
@frail jasper Has your question been resolved?
frail jasper
what would be the result if the expression is 5^k > 4^k+ 3^k + 2^k and you multiply 5 for each sides? would it be 5 * 5^k > 5 (4^k+ 3^k + 2^k)?
lone heartBOT
@frail jasper Has your question been resolved?
gusty gorge
what would be the result if the expression is 5^k > 4^k+ 3^k + 2^k and you multi...
yes that's multiplying 5 on both sides
jolly widget
Qn ii: which trig identity should I use?
lone heartBOT
Qn ii: which trig identity should I use?
jolly widget
Or am I perhaps approaching the question wrongly?
lone heartBOT
@jolly widget Has your question been resolved?
pseudo ice
Expand $R\sin(\theta + \alpha)$ with addition formula and equate coefficients
ocean sealBOT
@pseudo ice
jolly widget
Expand $R\sin(\theta + \alpha)$ with addition formula and equate coefficients
Qn is asking for me to simplify the expression to R sin(theta+a) tho, not the other way around 
And wdym addition formula?
pseudo ice
Write out, if you've seen it before:
[
R \sin(\theta + \alpha) = R \pqty{ \sin(\theta) \cos(\alpha) + \cos(\theta) \sin(\alpha) }
]
which you can then write as
[
\pqty{R \cos(\alpha)}\sin(\theta) + \pqty{ R \sin(\alpha)} \cos(\theta)
]
ocean sealBOT
@pseudo ice
pseudo ice
Better haha
ocean sealBOT
@pseudo ice
pseudo ice
Yea kinda
Therefore equating coefficients gives you that $R\cos(\alpha) = 8$, $R\sin(\alpha) = 6$ and from there you can find $R$ and $\alpha$
ocean sealBOT
@pseudo ice
main cedar
can someone please solve these?
lone heartBOT
can someone please solve these?
solemn juniper
How about you solve them and we can check your work
main cedar
How about you solve them and we can check your work
nyah im too embarrassed
solemn juniper
Well we're not going to do your homework for you
main cedar
Well we're not going to do your homework for you
lol, its not homewokr
solemn juniper
Ok
main cedar
ill try to do it the best i can
@solemn juniper I think my problem was I was mixing a-(2+a] and stupidly making a out of the bracket the coefficient
the ans is right
dunno about my method tho
any advice?
solemn juniper
a - 2a =/= 2
hot bluff
lone heartBOT
hot bluff
I started by finding the derivative of y
y' = 1/(x+1)^2
don't know how to proceed.
i tried just plugging in (1,2) to the slope intercept form but i got a tangent that didn't touch the curves.
or y - y_1 = m(x - x_1), forgot the name
frigid needle
once you get the derivative plug the X value from the point into the derivative equation. That answer will be the slope then use point slope formula to solve for the equation of the line
hot bluff
1/4, but i need other points that actually touch the curve
if i plug in (1, 2) i get a tangent that touches 0 points on the curve
frigid needle
could you take the derivative again and do the same thing?
jagged cobalt
note that (1,2) doesnt actually lie on the curve
hot bluff
correct, but how would i go about determining the points that i need
im used to having a point for the tangent, not a point that happens to intercept the tangent if thats the right wording
gusty gorge
basically, you have a point (x, y) on the curve
where y is given by the curve
now you need the slope of the line between (1, 2) and (x, y) to equal the derivative of the curve at that point
hot bluff
cause im blanking hard on how exactly id pick the two tangent points that happen to point to (1, 2)
gusty gorge
solve an equation to find them
hot bluff
yeah, but what equation?
i can guess points but id rather determine it algebraically if possible
jagged cobalt
$\frac{y_2-y_1}{x_2-x_1}=f'(x)$
ocean sealBOT
AℤØ
hot bluff
I already know the slope of (1, 2), = 1/4, and i know the derivative of y
jagged cobalt
no it doesnt
serene junco
(1,2) is not on the curve, so plugging it into the derivative of that function isn't really meaningful
jagged cobalt
now you need the slope of the line between (1, 2) and (x, y) to equal the deriva...
use this ^
hot bluff
(2 - y) / (1 - x)?
jagged cobalt
=f'(x), yeah
hot bluff
yeah but.... how does that help me. i know y' = 1/(x+1)^2
so wait
am i supposed to set y' = (2 - y) / (1 - x)?
so (2 - y) / (1 - x) = 1/(x+1)^2
serene junco
yeah, exactly. any point (x,y) on the curve that satisfies that will have a slope to (2,1) that's equal to the derivative at that point
in other words, it'll be tangent and pass thru (2,1)
hot bluff
so im on the right track by using that equation?
hot bluff
alright gimme a sec now
yeah idk still. i tried solving for y and got some weirdo answer. i tried checking my work and its showing completely different answers
i wounded up with
serene junco
I wouldn't try to solve for y
hot bluff
solving for x gets even more weird
serene junco
You have the equation for the curve y = x/(x+1)
substitute this for y
and then solve for x
since you're looking for a point on the curve, x and y must have the relationship that defines the curve
hot bluff
i gotta say, im still totally lost on this. what is the equation im supposed to be working with look like?
because i dont think (2 - y) / (1 - x) = 1/(x+1)^2 is the right one looking at the answer
serene junco
so `(2 - y) / (1 - x) = 1/(x+1)^2`
Does it make sense how you got this?
serene junco
because i dont think `(2 - y) / (1 - x) = 1/(x+1)^2` is the right one looking at...
it is
hot bluff
yeah its a relationship between the point (2, 1) and the derivative of our curve?
serene junco
Well, it says that the slope from (x,y) to the point (2,1) is equal to the derivative of the curve at (x,y)
which is true as long as the tangent at (x,y) goes thru (2,1)
hot bluff
derative is 1/4 at (x,y), but like we said earlier that point doesn't lie on the curve
serene junco
(x,y) is not (2,1)
(x,y) is a point on the curve
such that the slope between (x,y) and (2,1) is equal to the derivative at (x,y)
hot bluff
(2-x, y-1)
serene junco
I'm just trying to make sure the equation we're working with makes sense
Does it make sense why that equation has to be true for any point on the curve where the tangent goes thru (2,1)?
hot bluff
well, i know the derivative of a curve is the slope of the tangent at a point, and i know that we are trying to find that point in relation to (2, 1). but for some reason its just not clicking how i bring together the derivative of our curve when we dont even know the x or y
serene junco
For example, (0,0) is on the curve. But the slope from (0,0) to (2,1) is not equal to f'(0), so the tangent at (0,0) doesn't go thru (2,1)
we want the slope from (x,y) to (2,1) to equal f'(x), so that the tangent at (x,y) goes thru (2,1)
serene junco
because i dont think `(2 - y) / (1 - x) = 1/(x+1)^2` is the right one looking at...
which is what this equation says
slope to (2,1) = derivative
hot bluff
and in this case, our x is our input we are trying to find to plug into y'?
serene junco
well, our goal now is to solve for x, and that will be the x-value of the point on the graph that we're looking for
hot bluff
okay. let me have another go here
serene junco
make sure you substitute y=x/(x+1) first (or at some point while solving)
you're going to have a bad time if that equations still has two variables in it
ocean sealBOT
tatpoj
hot bluff
yeah that was the disconnect
serene junco
Notice I just substituted y with its definition: x/(x+1)
hot bluff
i was not doing that, i was simply leaving y in its place
serene junco
Gotcha, yeah we can do this because y=x/(x+1) is true for any point on the graph
hot bluff
right, its our y value at any x
hot bluff
im giving it a shot now
idk why i feel like im not doing this right
[2-(x/(x+1))]/(1-x) = 1/(x+1)^2 #join numerator fractions
(2x+2-x)/(x+1)(1-x) = 1/(x+1)^2 #(a/b)/c=a/bc
(x+1)^2[(2x+2-x)/(x+1)(1-x)] = 1 #???
Not exactly the route I took but I don't see a mistake
serene junco
**tatpoj**
it might be easier to just cross multiply from here
because notice when you multiply x/(x+1) by (x+1)^2, the denominator disappears
hot bluff
(x+1)^2[(x+2)/(x^2+1)] = 1
okay, sure
Yeah, this is fine but you turned a quadratic into a cubic
Leave the denominator in factored form instead
So you can cancel a factor of (x+1)
serene junco
```julia
(x+1)^2[(x+2)/(x^2+1)] = 1
```
oh and I don't think you meant (x^2 + 1) in the denominator exactly
but don't mind that, just leave the denominator factored in the first place
serene junco
um wait, hold up
serene junco
idk why i feel like im not doing this right
```julia
[2-(x/(x+1))]/(1-x) = 1/(x+...
from your last line here
see how you have (x+1) in the denominator?
and also you're multiplying by (x+1)^2
that's what I'm talking about
hot bluff
let me latex it
serene junco
should leave $\frac{(x+1)(x+2)}{1-x} = 1$
ocean sealBOT
tatpoj
serene junco
You had $(x+1)^2 \cdot \frac{x+2}{(x+1)(1-x)} = 1$
ocean sealBOT
tatpoj
hot bluff
$(x+1)^{2}(\frac{2-x}{1-x})$
ocean sealBOT
b0ngl0rd
serene junco
hm, I'm not really sure where 2-x came from, or how the x+1 in the denominator disappeared
serene junco
**tatpoj**
if you had this
serene junco
**tatpoj**
you just cancel a factor of (x+1) from numerator and denominator to get this
hot bluff
let me focus on the first fraction we deal with so we can do it in parts
2 - x/(x+1), first manipulation would be joining these?
serene junco
you could but I think there's a simpler way
Just cross multiply straight from the beginning
$(x+1)^2(2-\frac{x}{x+1}) = 1(1-x)$
ocean sealBOT
tatpoj
ocean sealBOT
tatpoj
ocean sealBOT
tatpoj
hot bluff
would you further expand after that
serene junco
yes, because now you want to solve for x
hot bluff
$x^{2}+4x+1=0$
ocean sealBOT
b0ngl0rd
hot bluff
is that anywhere close
hot bluff
what would be the best way to factor this and find the zeros
serene junco
It's not factorable
Just use the quadratic formula
or complete the square, i guess
hot bluff
is it weird for me to wanna complete the square here?
slate monolith
no
serene junco
No lol that's better
slate monolith
looks pretty easy yeah
serene junco
If you're comfortable with it, go for it
hot bluff
x^2 + 4x + (4/2)^2 + 1 - (4/2)^2 = 0
?
serene junco
sure
hot bluff
(x+2)^2-3 = 0
?
hot bluff
i feel like i cant see the forest for the trees im so deep into this problem, all the other problems weren't this long
so i got the above. does that mean we are looking at x = {2, -2}
serene junco
so `(2 - y) / (1 - x) = 1/(x+1)^2`
I think you might feel that way because you struggled a bit to solve the equation, but remember the point was to find values for x for which this is true
serene junco
so i got the above. does that mean we are looking at `x = {2, -2}`
no, if you add 3 to both sides, you'll get
(x+2)^2 = 3
so
x+2 = +/- sqrt(3)
so x = -2 + sqrt(3)
and -2 - sqrt(3)
hot bluff
duh, i had to isolate that factorization completely huh
serene junco
yeah but
serene junco
so x = -2 + sqrt(3)
and -2 - sqrt(3)
these are the x-values for the points on your curve where the tangent goes thru (1,2)
When you get stuck on a particular step for a while and end up spending a lot of time on the specifics, it's easy to lose track of the context
but just remember why you were solving that equation in the first place
hot bluff
hot bluff
that was just for x omggggg
serene junco
just plug the x-values into y=x/(x+1) to find the y's
personally I wouldn't bother but I guess that's what they want
serene junco
that was just for x omggggg
it's only "half" the answer, but you already did 99% of the work
in my experience, most calculus teachers don't care about things like simplifying radicals etc
hot bluff
he docked a pt because i didnt simplify [4/(-2a)]-1
hot bluff
thanks for bearing with me this whole time
idk why im struggling so hard on this problem, most of these were a breeze
serene junco
lol np
I didn't see the other questions, but if I had to guess, I think you might want to brush up on solving rational functions from algebra
You breezed through the actual calculus part of this problem but struggled with solving the equation
If the other problems were different types of functions, I can understand why they might have been a lot easier for you
hot bluff
i think it might be something even more simple like reducing complex fractions even
serene junco
that too, very similar concept
hot bluff
that and ive never thought to cross multiply, definitely wasnt my first instinct.
this problems given me a lot to think about
serene junco
ah yeah, whenever you have a/b = c/d, it's always a good choice
ad = bc
no more denominators
hot bluff
yeah thats way better than joining terms to the denominator
i also gotta be more careful on when i expand as a consequence of the manipulation i apply
serene junco
also, by making a common denominator, you would have had to multiply by (x+1)/(x+1), which actually introduces a new (fake) solution. Usually a good choice to avoid multiplying by a variable if possible
hot bluff
damn
serene junco
It would mean that you have to solve a cubic instead of a quadratic which is much more complicated
and one of the solutions would be extraneous anyway
hot bluff
well thanks again. 
hot bluff
it did, i guess its just one of those problems that shocks your system and makes you question your algebra instincts on how you could do it better. i felt like i was going a roundabout way
.close
stable night
lone heartBOT
stable night
hi can someone tell me how to start this qn? I have 0 clue as to how to start
tacit arch
Multiply top and bottom by the conjugate
stable night
stable night
Multiply top and bottom by the conjugate
Why?
tacit arch
Common technique for difference/sum of square roots
stable night
Common technique for difference/sum of square roots
i tried to simplify the top part but im not sure what to do to the bottom
square top and bottom?
tacit arch
You set a and b so that the limit is finite and 2021
lone heartBOT
lone heartBOT
Available help channel!
quartz panther
lone heartBOT
quartz panther
Im just so confused how to do this
vale wigeon
let A be a person's investment. write out what they are repaid in terms of A, p and q.
lone heartBOT
@quartz panther Has your question been resolved?
west cipher
divide by 0
you must factor
what did you factor
you factored to find x, right?
so show my
what equation did yo factor
alpine sable
use the quadratic equation if you want
alpine sable
or find two numbers that multiply and give -2
but when added give +3
west cipher
do you know how to factor
ok
if you want, you can do this
divide both sides by two
thus
alpine sable
no
west cipher
now you can factor it if you are familiar with that form
ocean sealBOT
alpine sable
eeez
you need the quadratic formula
for this one
no other way
you've copied the question wrong then
if you're in 8th grade
nah try again
send a ss
it's meant to be 2x^2 + 3x +1 =0
from your book
vague mirage
Use Quadratic
ocean sealBOT
vague mirage
$x=\frac{-b±\sqrt{b²-4ac}}{2a}$
alpine sable
dw about it, you'll learn it later
ocean sealBOT
Offline due to exams
vague mirage
I'm in 9th grade
lone heartBOT
@hushed ether Has your question been resolved?
mint lodge
lone heartBOT
subtle birch
of course
grave cedar
so cb should be 9 cm right?
mint lodge
no it should not
grave cedar
why so
mint lodge
two different triangles can have same area without
grave cedar
two different triangles can have same area without
but 2 sides are same
mint lodge
without same sides
grave cedar
u see?
grave cedar
two different triangles can have same area without
cm is common am=bm
mint lodge
yes
grave cedar
CM is for both
yes
mint lodge
ok then what
mystic elk
does Z_17 have any non trivial sub group ?
lone heartBOT
does Z_17 have any non trivial sub group ?
desert zephyr
Try looking at the orders I think
mystic elk
how ?
mystic elk
17
desert zephyr
And what do you know about the order of its possible subgroups?
desert zephyr
Lagrange's Theorem
From there you should be able to conclude that the only subgroups are trivial or whole group
mystic elk
but Lagrange theorem is for cosets and index right ?
desert zephyr
You can read up on it on Wikipedia. It's used for the index yes but it also follows that if there's a subgroup the order s should divide
ancient ivy
lone heartBOT
lone heartBOT
@ancient ivy Has your question been resolved?
lone heartBOT
@ancient ivy Has your question been resolved?
lone heartBOT
@ancient ivy Has your question been resolved?
sour breach
Hint:
||
Let x = sqrt(111...111), then we can write:
x^2 = 111...111
10x^2 = 1111...1110
Subtracting the two equations yields:
9x^2 = 1111...1101
Dividing by 9:
x^2 = 1234...5678.9
||
lone heartBOT
@ancient ivy Has your question been resolved?
ancient ivy
Hint:
||
Let x = sqrt(111...111), then we can write:
x^2 = 111...111
10x^2 = 1...
How can i know the position n+1 after x
ancient ivy
.close
outer spear
8 x 7
outer spear
sorry i just have bad maths
mortal trellis
use a calculator
vague mirage
sorry i just have bad maths
What's your age if you don't know 8×7
And you don't have any more questions right?
alpine sable
.close
celest acorn
Question 2 please help
lone heartBOT
Question 2 please help
alpine sable
send a clear picture
celest acorn
Its clear
alpine sable
if the person who is trying to help you tells you it isn't clear
and you tell them it's clear
celest acorn
alpine sable
how do you expect to receive help
alpine sable
well look at the boundaries
celest acorn
What are they?
alpine sable
look at the boundaries for the lenght
celest acorn
It goes to the right.
I'm not even trolling bro I'm confused.
Like really.
Can you do a worked example?
@alpine sable
lone heartBOT
@celest acorn Has your question been resolved?
lime wyvern
hybrid ether
lone heartBOT
tacit arch
Are you given OTP is a right angle?
hybrid ether
Nope
modern sedge
I guess it's tangent?
hybrid ether
Yes
modern sedge
Do you know a famous formula that relates sides of right triangle?
usually a, b and c
modern sedge
yep
hybrid ether
Thanks
modern sedge
Dont advertise here please
hybrid ether
Is this q same?
modern sedge
Is this q same?
not same, rather similar
misty bobcat
Given a binary relation $R$ over $Z_n$, decide if there are 8 relations
R1, R2, R3, R4, R5, R6, R7, R8 such that R is their union and all relations satisfy:
$\forall x \forall y \forall z (Ri(x,y) \wedge Ri(x,z) \implies Ri(y,z))$
ocean sealBOT
Michal
misty bobcat
does somebody know how to express this in CNF (conjunctive normal form) ?
vale wigeon
youve been posting this problem in the exact same form for several days in a row now and have gotten, from what i've seen, zero response.
no, get your own. #❓how-to-get-help for instructions.
if you're looking to express this in CNF, you're gonna want a bunch of binary decision variables.
i assume Z_n here just means {0, 1, ..., n-1}?
it feels like the most natural choice of variables is $x_{i,j}^k$ where $i, j \in Z_n$, $k \in 1:8$ and the meaning of $x_{i,j}^k$ is ``the pair $(i,j)$ belongs to relation $R_k$''.
ocean sealBOT
Ann
vale wigeon
for a total of 8n^2 variables.
then the condition that $R$ is the union of your relations is encoded by adding, for every pair $(i,j) \in R$, the clause $x_{i,j}^1 \lor x_{i,j}^2 \lor \dots \lor x_{i,j}^8$.
ocean sealBOT
Ann
vale wigeon
and the other condition is... i guess encoded with the clauses $\neg (x_{i,j}^k \land x_{i,l}^k) \lor x_{j,l}^k)$?
ocean sealBOT
Ann
vale wigeon
which would be $\neg x_{i,j}^k \lor \neg x_{i,l}^k \lor x_{j,l}^k$
ocean sealBOT
Ann
vale wigeon
i suppose
misty bobcat
i assume Z_n here just means {0, 1, ..., n-1}?
yes, exactly
misty bobcat
it feels like the most natural choice of variables is $x_{i,j}^k$ where $i, j \i...
i managed to do something like this
vale wigeon
were you asked to do anything beyond encoding this as a SAT problem
misty bobcat
which would be $\neg x_{i,j}^k \lor \neg x_{i,l}^k \lor x_{j,l}^k$
I did exactly this
vale wigeon
were you asked to do anything beyond encoding this as a SAT problem
misty bobcat
were you asked to do anything beyond encoding this as a SAT problem
i need to put it correctly into clauses and put the output into sat solver
vale wigeon
,w conjunctive normal form
ocean sealBOT
Requested by Ann#0413
vale wigeon
fuck off...
okay, off to wikipedia we go
ok, CNF means "an AND of ORs"
so the thing i wrote would appear to be correct.
did it (a) get accepted, (b) get rejected on a technicality, (c) get rejected for something meaningful, or (d) not get submitted at all?
misty bobcat
up to now i coded it up, now im about to test it
here is my code so far
Cnf decomposeRelationToCNFNonBinary(const Relation& r, unsigned int n){
Cnf cnf;
// // \forall x \forall y \forall z (Ri(x,y) & Ri(x,z) -> Ri(y,z))
// A & B => C <=> ¬A ∨ ¬B ∨ C
for(int k = 0; k < 8; k++){
new_clause(cnf);
for(auto re1: r){
for(auto re2 : r){
// re1 & re2 => re1.second == re2.second
write_literal(cnf, -s(re1.first, re1.second, k, n));
write_literal(cnf, -s(re2.first, re2.second, k, n));
write_literal(cnf, s(re1.second, re2.second, k, n));
}
}
}
// each pair must be in one relation R_k at least
for (auto re: r){
new_clause(cnf);
for(int k = 0; k < 8; k++){
write_literal(cnf, s(re.first, re.second, k, n));
}
}
for(auto& clause: cnf){
for(auto literal: clause){
cout << literal << " ";
}
cout << endl;
}
return cnf;
}
oof
alpine sable
**Michal**
what the flip is this
misty bobcat
what the flip is this
flip? dont get it
vale wigeon
"flip" in this context is a censored version of "fuck"
lone heartBOT
@misty bobcat Has your question been resolved?
vale wigeon
and how is that indicative of a mistake...?
misty bobcat
i made a mistake probably, now im going to try to follow exactly your solution
I coded algorithm which do the decomposition, also i used big input
misty bobcat
which would be $\neg x_{i,j}^k \lor \neg x_{i,l}^k \lor x_{j,l}^k$
each condition like that should be in new clause?
this is my input and it's satisfiable, i believe that this is NOT satisfiable {{0,0}, {0,1}, {0,2}, {0,3},{0,4}, {0,5}, {0,6}, {0,7}, {0,8} }
vale wigeon
each condition like that should be in new clause?
yeah
im not sure i can debug your code
misty bobcat
im not sure i can debug your code
you do not have to, i will do it myself, i just need and ideas and guidance 🙂
dont get it, i think that it's correct translation of your ideas
Cnf decomposeRelationToCNFNonBinary(const Relation& r, unsigned int n){
Cnf cnf;
for(int k = 1; k <= 8; k++){
for(auto re1: r){
for(auto re2 : r){
// re1 & re2 => re1.second == re2.second
// // \forall x \forall y \forall z (Ri(x,y) & Ri(x,z) -> Ri(y,z))
// A & B => C <=> ¬A ∨ ¬B ∨ C
new_clause(cnf);
write_literal(cnf, -s(re1.first, re1.second, k, n));
write_literal(cnf, -s(re2.first, re2.second, k, n));
write_literal(cnf, s(re1.second, re2.second, k, n));
}
}
}
for (auto re: r){
new_clause(cnf);
// each relation element must be at least in one of the 8 relations (union condition)
for(int k = 1; k <= 8; k++){
write_literal(cnf, s(re.first, re.second, k, n));
}
}
// Open the file for writing
std::ofstream outfile("data.txt");
// Write the vector to the file
for (const auto& row : cnf) {
for (const auto& element : row) {
outfile << element << " ";
}
outfile << "0\n";
}
// Close the file
outfile.close();
return cnf;
}
@misty bobcat Has your question been resolved?
lone heartBOT
@misty bobcat Has your question been resolved?
lone heartBOT
@misty bobcat Has your question been resolved?
mellow mural
can someone please help me with this: I am just trying to find side length z and nothing else:
lone heartBOT
can someone please help me with this: I am just trying to find side length z and...
@mellow mural Has your question been resolved?
steady spoke
lone heartBOT
steady spoke
can anyone show me step by step on how to solve this
outer lark
I would start by splitting it up into shapes you know how to compute the area of
steady spoke
I would start by splitting it up into shapes you know how to compute the area of
this is like a practice question i found online and i just wanted to see an example of how someone would solve it
toxic moth
all the pieces are rectangles or right triangles, which are half rectangles
you can also use pick's theorem
steady spoke
you can also use pick's theorem
yeah but can you show me how i can do it i have a similar problem like this
and i wanna use this as an example
toxic moth
you know how area can be gotten with length times width right?
steady spoke
yes
toxic moth
ill draw something real quick
only the top left square and top middle rectangle are fully in the polygon of course
but you could get the area of these six and then its pretty obvious to figure out from there
steady spoke
you wouldnt have one trapezoid and one triangle?
toxic moth
you mean like splitting with a line from B to D?
you could do it that way, but then the right side triangle is still easily done like how i drew, since its not a right triangle
and then split up the trapezoid in two
west girder
you can use picks or shoelace
lone heartBOT
@steady spoke Has your question been resolved?
steady spoke
you mean like splitting with a line from B to D?
how abt from B to E
well nah it would be the same ig
toxic moth
well theres an extra small bit to it that im hoping you've figured out
like for the red one you cant take the area of the entire outer square
steady spoke
like for the red one you cant take the area of the entire outer square
i mean i was hoping you could show me step by step because this isnt my question. i jsut wanted someone to follow
something*
😭
toxic moth
for the red version, 3 out of 4 of them are split in half, while one is fully part of the polygon
steady spoke
this is my real one
toxic moth
looks pretty similar
toxic moth
but anyway
if you do length times width for those 3 pieces what do you think happens?
if half of them is in the polygon we need, and half out
lone heartBOT
@steady spoke Has your question been resolved?
alpine sable
how do i do this?
lone heartBOT
how do i do this?
ornate condor
um
if its
a+b sqrtc
- top n bottom by a-b sqrtc
if sqrta+b sqrtc
- sqrta -b sqrtc
if its (stuff)^2
expand it first
serene junco
sure
serene junco
what did you end up with?
alpine sable
what did you end up with?
19 + 8√3
serene junco
multiply the numerator and denominator by the conjugate: 19 - 8sqrt3
alpine sable
ok
serene junco
It should leave you with an integer in the denominator
serene junco
yeah, then just clean up the numerator
alpine sable
oh ye, i got it. Thanks a ton
alpine sable
.close
molten mountain
Hello
molten mountain
I need helop on this question
I would like you to help explain how you do this problem
thank you
lone heartBOT
@molten mountain Has your question been resolved?
lone heartBOT
@molten mountain Has your question been resolved?
alpine sable
$\frac{\sqrt[3]{27x}}{\sqrt{x}}=\frac{\sqrt[3]{27}\sqrt[3]{x}}{\sqrt{x}}=\frac{3\sqrt[3]{x}}{\sqrt{x}}=\frac{3x^\frac{1}{3}}{x^\frac{1}{2}}=x^\frac{3}{6}-x^\frac{2}{6}=x^\frac{1}{6}=\frac{3}{x^\frac{1}{6}}$
ocean sealBOT
okokok
alpine sable
simplifying
median oar
.close
lone heartBOT
molten pivot
$\frac{\sqrt[3]{27x}}{\sqrt{x}}=\frac{\sqrt[3]{27}\sqrt[3]{x}}{\sqrt{x}}=\frac{3...
False
You forgot to write the 3 in the 5th and 6th expressions
fierce prairie
it's being dealt with in #help-7｜zen1thxyz
lone heartBOT
Available help channel!
lapis tartan
Hey, I got help getting started with this question earlier and this is my best attempt at it. I’m not totally sure I’m correct, and am hoping someone can spot where I go wrong here. Thanks!
vague mirage
Why the answer is negative?
lapis tartan
well i thought it would be q3 because the question says cosine is negative, and tan is positive
strange meadow
Yes it’s q3 good
lapis tartan
Why the answer is negative?
originates from my first declaration of cos(theta), where i make it -7/sqrt(58)
vague mirage
well i thought it would be q3 because the question says cosine is negative, and ...
Then yes quadrant 3
Sine is negative
strange meadow
^
vague mirage
But it's sin 2theta
lapis tartan
oh yea sine is negative in q3
ocean sealBOT
vale wigeon
Hey, I got help getting started with this question earlier and this is my best a...
you dont need to know theta itself for this question
actually i dont think you even need to know the quadrant
the value of tan(θ) is enough to find sin(2θ)
vague mirage
Ok $\sin(2x)=2\sin(x).\cos(x)$
vale wigeon
\cdot
lapis tartan
i dont follow
ocean sealBOT
𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛
vale wigeon
after all, $\sin(2\theta) = 2\sin(\theta) \cos(\theta) = 2 \tan(\theta) \cos^2(\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)}$
vague mirage
ocean sealBOT
Ann
vague mirage
You can use any of em
vale wigeon
transparent background yikes
vague mirage
transparent background yikes
I didn't get a perfect background pic 🥲
lapis tartan
my answer could technically be sin2(theta) = 2tan(3/7)/1+tan^2(3/7) ?
vale wigeon
no.
lapis tartan
sorry... dont know the formatting to make it fancy
vale wigeon
it is not the formatting that is the issue
vague mirage
Tan theta =3/7
lapis tartan
im kinda getting lost, is it possible to help me understand where i went wrong?
vague mirage
$\frac{2×\frac{3}{7}}{1+(\frac{3}{7})²}$
lapis tartan
or was i never on the right track?
ocean sealBOT
𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛
vague mirage
Yupp now good
vague mirage
or was i never on the right track?
You just had to put the value of tan in
vale wigeon
im kinda getting lost, is it possible to help me understand where i went wrong?
we are suggesting a different route to you than you showed on paper.
vague mirage
You just had to put the value of tan in
But what you did was you are making is theta = 3/7
vale wigeon
we sidestep the calculation of sin(1θ) and cos(1θ)
lapis tartan
i understand
lapis tartan
but to be clear that would be my answer?
and thank you both for the guidance in moving to what is very much a simpler identity to use in this scenario
vale wigeon
sure would
lapis tartan
.close
median oar
Why does this book say rational
fallen verge
irrationals go wack
still possible, but limit definition is probably not the way to go
fallen verge
show me using limit def how to do it
pseudo ice
Well I guess because of the way that they showed it, in the steps below?
fallen verge
oh fair the proof probably works for rationals because but p and q in p/q need to be integers
echo moth
lone heartBOT
echo moth
I just need some guidance on which steps to take
slate monolith
I just need some guidance on which steps to take
what have you tried so far
echo moth
what have you tried so far
I have this so far but I don’t know if it’s the right direction