why is it 2sin(a) cos(b)
#help-28
sour pine
I don't get your question
winter escarp
wait
wait
wait
am i being dumb
so
cos(b) + cos(b)
is 2cos(b)
so why
is sin(a)cos(b) + sin(a)cos(b)
only
2sin(a)
cold agate
Expand sin(a+b) and sin(a-b)
sour pine
The hell
winter escarp
is it bc its the multiplying factor
sour pine
I don't get what you mean @winter escarp
cold agate
If you forgot the formula, you can derive it via multiplying out e^ix and e^iy
sour pine
yeah
sour pine
or just some coordinate geometry although complex numbers approach would be best
winter escarp
why doesnt it become 2cos(b)
if u are adding
cos(b) together
like surely
if u doing
sin(a)cos(b) + sin(a)cos(b)
thats =
to
2(sin(a)cos(b))
sour pine
@cold agate do u get what he means
cold agate
2(sin(a)cos(b))
Yeah that's the identity
cold agate
what
cold agate
Let's take a step back
winter escarp
lets say
cold agate
And start from the start
cold agate
.
winter escarp
but i dont get why it isnt 2cos(b)
bc we are adding 2 cos(b)'s together
😭 IM HELPLESS
cold agate
bc we are adding 2 cos(b)'s together
We're not
We're adding sin(a)cos(b) to sin(a)cos(b)
cold agate
That is the identity
winter escarp
uni stress
is like
making me back
imma implode
im too far gone
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cerulean tangle
how to calculate meteorite original mass before being reduced to 20 grams at a speed of 11 km/s
cerulean tangle
assume that ablation is 5x10^-8 J/kg
AI gave me an answer but it can't be trusted
brittle parcel
AI gave me an answer but it can't be trusted
did it give you a suggested diffeq family
brittle parcel
ah i don't actually know enough about how to model this to help
teal basalt
|x|
absolute
is when you make any value into a positive value.
is there a reverse one, that make the value into negative?
teal basalt
like what is the name?
sudden condor
-|x|
teal basalt
hmm...
brittle parcel
there isnt a standard name because the point of absoljute value is that it tells you the number's "size"
and that's inherently a non-negative concept... usually
teal basalt
I see... thx all
green merlin
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teal basalt
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rapid nymph
Hey guys, what’s the equation for how many digits are in a number
slate violet
Hey guys, what’s the equation for how many digits are in a number
do you know what logarithms are?
rapid nymph
nah fam
slate violet
well you need them for the equation
say if we are using the base 10 logarithm, there's this nice pattern
log(10) = 1
log(100) = 2
log(1000) = 3
log(10000) = 4
and so on
rapid nymph
Why ?
onyx glen
like not just for the specific matter of figuring out a number's digit count but also because you most likely will see logarithms as you go further into math and/or computer science
so like
you need to know that shit
slate violet
it's a general rule in every subject that to answer more complicated questions, you need more complicated concepts
rapid nymph
log(10) = 1
log(100) = 2
log(1000) = 3
log(10000) = 4
and so on
How can I make it? 0.01 =1
slate violet
log(10) = 1
log(100) = 2
log(1000) = 3
log(10000) = 4
and so on
logarithms are the thing that let you do this
slate violet
How can I make it? 0.01 =1
oh so you're asking about decimals?
rapid nymph
So basically the difference between calculators and computers
slate violet
like how many digits 0.023 has?
rapid nymph
Yeah fam
rapid nymph
No I’m trying to make a calculator
slate violet
I see
rapid nymph
It’s a little bit stupid, but I can show you it
slate violet
sure actually
slate violet
like how many digits 0.023 has?
honestly for this it's just the number of digits after the decimal point, after you remove the trailing zeroes
you could easily convert 0.023 to a string and get the string length
rapid nymph
sure actually
slate violet
you could easily convert 0.023 to a string and get the string length
cause the log approach doesn't work with decimals
analog sand
i think modulo operator is a good alternative
slate violet
how so
slate violet
I know but how exactly would you use modulo
analog sand
since for any number, N% 10^x belongs to [0,9] would give you the digits x
rapid nymph
cause the log approach doesn't work with decimals
String?
I’ve made eight functioning calculator
analog sand
Yeah, I did code
which lang
analog sand
this isnt code
rapid nymph
I thought it was
analog sand
coding is using a coding language
rapid nymph
I what’s your definition of coding
analog sand
like python or cpp
rapid nymph
I doubt coding with anything that can do with advanced math, formulas, and hold variables
bright bronze
tbh desmos is turing complete, but it is unwieldy
analog sand
didnt know that
onyx glen
the 10 in log_10 should be a subscript or gone entirely
onyx glen
tbh desmos *is* turing complete, but it is unwieldy
i told op to learn a proper programming language but he apparently has no computer and only a phone
ruby agate
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ruby agate
Solution any knowers plox?
stiff musk
shoot it and move on
foggy vapor
Mfw roblox
foggy vapor
Wtf are the floating minus signs meant to represent
shy bane
Bro what are u playing 💀
ruby agate
Double negative if you mean - -30e^pi
foggy vapor
No I mean what is this 
ruby agate
Double negative
foggy vapor
kinda weird but ok
analog sand
ah yes -+
proven frost
evaluate each term step by step
stiff musk
Wtf are the floating minus signs meant to represent
that's the weapon crosshair
shy bane
There are many double negatives
ruby agate
that's the weapon crosshair
Oh shit
stiff musk
notice there are vertical ones too
ruby agate
Oh that one in particular is but there are double negatives
foggy vapor
Anyways you can work it out yourself. These aren't actually impossible to solve; just long
stiff musk
yea someone's being a clown here
analog sand
first part is just 7^3 + ( 8^2+13)
foggy vapor
Not sure if !nosols'ing a roblox game is appropriate but aye
ruby agate
The first equation entirely or the first parenthesis
analog sand
Not sure if !nosols'ing a roblox game is appropriate but aye
lmao
analog sand
The first equation entirely or the first parenthesis
everything above the 'and'
ruby agate
Cool ty now what ab after
shy bane
Maybe you don't have to solve it it's literally a game check for clues
analog sand
do that yourself lmao i aint gonna sit and do sigma for roblox
shy bane
Lmao
foggy vapor
Watch the answer be 69 or some haha funi number
ruby agate
Im not the one playing it my friend just asked if anyone could solve so i brought it here
shy bane
Check for gameplay in yt
shy bane
Watch the answer be 69 or some haha funi number
It's five digit do 69696
ruby agate
Check for gameplay in yt
Idk what game it is
shy bane
Idk what game it is
Then why u playing
analog sand
69420
shy bane
Or solving
analog sand
more likely
ruby agate
Then why u playing
IM NOT i just said that
shy bane
IM NOT i just said that
Eh?
foggy vapor
Why are u being the middleman to a roblox game 
shy bane
Literally
shy bane
Lol
ruby agate
A man cant be invested?
analog sand
this is definitely a trick lol, the first parentheses below the and doesn't even work out to be an integer
its e^i pi format
shy bane
Tell him to find clues
ruby agate
Oh true ok
shy bane
Around the room
ruby agate
Thanks
analog sand
ruby agate
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maiden thorn
base 10
log(ax) = 2log(x + 1)
- Is it true that for only a = 4 the equation has integer root?
Don't know how to solve 1.
- Is it true that the equation has 2 root only for a > 4?
So, I started solving and got (-inf; 0) & (4; inf) for a. But the solution is (4; inf), which means 2. is true, but I don't fully understand how to "eliminate" (-inf; 0)
cyan walrus
uh
idk what you tried to do with intervals
ax > 0
x + 1 > 0
these conditions must always be true
now,
(x+1)² = ax
a = x + 1/x + 2
a is an integer for x = 1 and -1,
now test for which values of x and a, the inequalities hold true, and that should be a sufficient proof
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maiden thorn
now test for which values of x and a, the inequalities hold true, and that should be a sufficient proof
but how does it help me with a > 4. I wanted to know where the equation will have 2 solutions.
- ax = (x+1)^2
- ax = x^2 +2x + 1
3.x^2 + (2 - a)x + 1 = 0
D = (2-a)^2 - 4 should be > 0 to have 2 root
D = 4 - 4a + a^2 - 4 > 0
a^2 - 4a > 0
a = 0, a = 4 => (-inf; 0) & (4; inf) for a
Since ax > 0 and x > -1 =>
either a is (-inf; 0), x is (-1; 0) or a is (0; inf), x is (0; inf)
What am I doing wrong?
maiden thorn
Yes, that is what confuses me
cyan walrus
a has to be an integer no?
cyan walrus
it's not stated,
but if a is any real number, there are infinitely many integer roots..
maiden thorn
It doesn't say int root for a > 4, it says does the equation have 2 roots only for a > 4
maiden thorn
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weary vector
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weary vector
I am stuck here if someone can help me.
|x+y-5| if x+y can be -4 and it also can be 5 is this positive or negative? because i get values for both
i get 0 and -9
and of course what other values we get in between for [-4;5]
rocky rapids
!xy
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dull flint
which formula is used here
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which formula is used here
dull flint
which one Tn = ar^n-1 or Sn = a(r^n-1)/r - 1
fast peak
for starters its an infinite series
so neither of those
1+r+r^2+... = 1/(1-r)
or rather a+ar+ar^2+... = a/(1-r)
dull flint
S = a(1 - infinite)/1-r ??
onyx glen
or rather a+ar+ar^2+... = a/(1-r)
1-r not 1+r
fast peak
whoops thanks
digital hemlock
(r^(n+1)-1) is zero for r=1, so has a factor (r-1), the division gives (r^n+...+r+1)
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final wing
Contradiction, Assume that no vertex has in degree 0. Now we can construct a directed closed walk : v0,a1,v1,a2,v2,....,a_k-1, a_k, v0 . THis is a contradiction so D has at least one vertex with in degree 0.
final wing
I know some vertices can have out degree 0 but there still must be a loop cuz otherwise some will have in degree 0 which we just assumed is not possible
jolly flare
I know some vertices can have out degree 0 but there still must be a loop cuz ot...
you can prove the contrapositive p easily here I think
final wing
what?
jolly flare
what?
like prove if D does not have at least one vertex with in-degree 0, then it must contain normal cycle(s)
final wing
isnt that kind of what I did?
jolly flare
isnt that kind of what I did?
oh yeah you did to it by contraposition
mb i didnt see
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final wing
<@&286206848099549185>
nimble ridge
You have the right idea, but you have to write it down correctly. Start wit a random point v0. THe edge leading to v0 has to start at another point v1.
If we have a set of connected vertices {v0; v1; ... ; vk} the last one has in-degree 1, where does th edge leading to this one start? it has to be a new point v_{k+1}.
final wing
I thought v0 since v0 otherwise would have in degree 0
nimble ridge
Any of the points we already used, but it does not matter, it will be a loop.
final wing
True but then we dont use that all vertices have at least in degree > 0 right?
nimble ridge
If they had no in-degree >0, how could I argue, that there is an edge leading to vk?
final wing
Could you maybe rephrase that for me
nimble ridge
I start at the end of a chain and state, that all vertices have in-degree >0. Then there must be an edge leading to each vertex.
final wing
true
nimble ridge
If they could have in-degree 0, we could reach a point, where we can not continue the chain.
final wing
exactly
nimble ridge
SO the chain continues until we reach an already used point (loop) or the last element of V.
But that last point also has in-degree >= 1 and we have to use one of the previously used vertices.
final wing
but what about the first point of our chain?
nimble ridge
Think of it as a lasso, it does not have to connect to the first point, v0 could have out-degree 0.
e.g. v0 <- v1 <- v2 <- v3 <- v4 <- v2, where v2 has out-degree 2
final wing
But now v2 doesnt have an in degree greater than 0 right?
nimble ridge
No, but if the in-degree is at least 1 and we have a vertex with out-degree 0 , at least one vertex has to have out-degree >= 2.
final wing
I am tweaking :-:
nimble ridge
Sorrry, v2 has an in-degree 1, edge from v3 to v2.
nimble ridge
yes!
final wing
but how do I write this down nicely?
nimble ridge
I would write it as a sort of induction, start with 2 vertices and then show k->k+1 for some k < |V|
nimble ridge
Sorry, but I think you should try this yourself first. You got the general idea.
YOu do not have to write it as a formal induction, but the steps are similar.
final wing
Hmmmm
i am stuck
Why do I not get itttttttt
Omg i got it
not with induction
but I think how I wrote the proof is also fine
We have a graph with |V| = n. We assume by contradiction that every vertex has an in-degree >0. We choose a random vertex v in our graph. We know this vertex has an in degree greater than 0 so we can go to the vertex that points to our randomly picked vertex. We can keep doing this. After max n+1 times we will have arrived at a vertex we already passed before. Hence there exists a loop => contradiction
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rare pine
Let ABC be a triangle where AB = sqrt(3) and D is a point on BC such that AD = 1. Find the value of BD*DC
rare pine
theres probably some theorem for this but i forgot so im trig bashing
is there any way to trig bash the solution? cause i cant find it
onyx glen
Let ABC be a triangle where AB = sqrt(3) and D is a point on BC such that AD = 1...
got a diagram?
onyx glen
wow ok this is ugly
rare pine
following this from the second img i have that k = 3 which is ridiculous
<@&286206848099549185>
blehh
glad haven
sexiest writing ive ever seen.
cold agate
Such nice handwriting
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rare pine
Redo and still stuck.
torn jolt
What’s the prob though ?
rare pine
Let ABC be a triangle where AB = sqrt(3) and D is a point on BC such that AD = 1. Find the value of BD*DC
rare pine
why am i trig bashing this qn for no reason💔
torn jolt
Product of lengths bd and dc ?
hushed barn
Let ABC be a triangle where AB = sqrt(3) and D is a point on BC such that AD = 1...
Uh what. I’m pretty sure there’s not enough given information. Like fro example in the diagram you can extend BC such that BD will always remain the same and CD can be variable
hushed barn
Uh what. I’m pretty sure there’s not enough given information. Like fro example ...
And triangle BDA remains same. It’s just moving point C along the line BC
torn jolt
rare pine
Uh what. I’m pretty sure there’s not enough given information. Like fro example ...
I KNEW ITTTT
something was fishy about this one
torn jolt
Is the answer 4 though ?
rare pine
i honestly have no idea
my friend gave me this for me to solve since im good at math n allat
torn jolt
This question’s incomplete maybe
rare pine
ill ask him !!!
torn jolt
Sure
rare pine
This question’s incomplete maybe
yea probably mistyping
torn jolt
Nice handwriting though
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rare pine
Is my solution correct?
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hollow hemlock
Optimization Problem:
Is my answer correct and (my own question) how would I go about solving it - given there are two variables to optimize for.
On a 1000x1000 grid two colleges are planned A and B as well as 1000 students. A can handle 400 students. B can handle 600. What's the optimized location for A and B so that the total sum of distance (college and student) is minimized.
I'd write something like:
Required sum(1 to 1000) z_i = 400 //Students of school A
z_i either {0;1} for 0<i<=1000 //if Student goes to A set 1. Else 0.
variables are real numbers```@hollow hemlock Has your question been resolved?
hollow hemlock
<@&286206848099549185>
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ocean steeple
hi I am doing pre-cal and I know we are supposed to use cos,sin or tan but the answer does not make sence to me I keep getting different numbers which the answer is 5.638
ocean steeple
hard phoenix
What were your steps?
ocean steeple
Determine the variable x. and round your answer to the nearest thousandths.
hard phoenix
But what did you do to try and solve it?
ocean steeple
I tried tan since it was x and y but I got 2.93
hard phoenix
Uh
Ok so tangent of angle gives opp/adj right?
So in this case the side opposite to the 70 degrees angle is x
But you don’t know the adjacent side
So you can’t really use tan here
ocean steeple
can you writre it down what your saying sorry I learn when it is written down
hard phoenix
Sorry that’s not really possible for me rn
ocean steeple
ahhh okay that's fine then
limpid kiln
For this case you shall want to use sine since you have the opposite side and the hypotenuse to work with
ocean steeple
ahhh okay so would my signs switch and my x be x and my y be 6?
limpid kiln
Yes indeed
limpid kiln
Sine is opposite divided by hypotenuse so you’ll want the fraction the other way
H/6
limpid kiln
Correct
ocean steeple
so what is the next step?
limpid kiln
Isolate H
ocean steeple
do I times H in both sides?
limpid kiln
Multiply by 6 to cancel out with the denominator
ocean steeple
limpid kiln
Yes
ocean steeple
On the answer key yes but no when I’m solving it on the answer key it says x=5.638
limpid kiln
If you shall desire an approximate result use a calculator
To find the decimal value of 6*sin(70)
ocean steeple
I don’t have a calculator on me will the computer help me?
limpid kiln
Well there are multiple resources you can use on a computer
You may already have a calculator app
Now unfortunately I have to go, good luck
fading stone
Differential Equations:
I've attempted variations of this problem like 5 times and I think I'm doing the same thing wrong each time, but I can't pinpoint where my mistake is.
Call the first term of the equation Mdx, and the second Ndy.
Call the solution of the equation F(x,y)
Call (Mx - Nx)/M) = P
The equation is not exact, but P = (Mx - Nx)/M) is a function of just Y, so I can use an integrating factor mu(y) - exp[int((Mx - Nx)/M)dy].
My understanding is that del(F)/del(x) = M and del(F)/del(y) = N.
F(x,y) = int(Mdx) + g(x)
Using that information and the integrating factor, I can substitute as necessary, but I'm doing something wrong or working off of false logic, and I can't figure out what I'm doing. I'll attach photos, for context, of the prompt from my homework as well as a photo of my work for my most recent attempt.
fading stone
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nimble crane
I can confirm that your integrating factor looks good
you've turned the ODE into an exact ODE
nimble crane
okay, I think I see the problem
you wrote that -3x^2y^-4 + g(y) = 2y - 3x^2
this isn't true
when you integrate M = 2xy^-3 with respect to x, you want to get a function F that satisfies F_x = M, F_y = N, where N = y^-4(2y - 3x^2)
the F needs to satisfy those conditions for the exact ODE, not the original ODE
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fading stone
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nimble crane
so after you integrate M, you're supposed to get x^2y^-3 + g(y) = F(x, y)
differentiate wrt y to get -3x^2y^-4 + g'(y) = N = 2y^-3 - 3x^2y^-4
and from there you can identify g'(y) with 2y^-3, and be on your way 
idk if the way I said this makes sense 
fading stone
the F needs to satisfy those conditions for the **exact** ODE, not the original ...
so I need to set F(x,y) = int(M mu dx) as opposed to int(M dx)?
F(x,y) = the integral of the first term TIMES the integrating factor wrtx?
if im understand correctly it looks like ive done that
nimble crane
so I need to set F(x,y) = int(M mu dx) as opposed to int(M dx)?
yes
you did
this is the part I'm concerned about
nimble crane
you set the derivative of F to be equal to the "N" in the original ODE
not the exact ODE
nimble crane
yes
nimble crane
that should solve the issue
fading stone
while you're here, do you think you could help me understand this solution?
I understand that I'm using an integrating factor to turn the differential equation into an exact differential equation and solving from there. I think I mostly understand the solution of the exact differential equation but I could not articulate back to you why del(f)/del(y) = n (or apparently n*mu).
Integrating factors also still sorta feel like magic. I see that they provide enough new information to make the equation solvable and of course if you multiply both sides of the equation by the same factor then it all remains equal, but I don't understand how exponentiating the integral of some term or partial derivative from the function creates such a useful integrating factor
does that make sense?
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idle orchid
Hello does someone know calculus, derivative to be exact i have an exam tomorrow and i need some help
stiff musk
what is your question?
slender bane
what concepts do you understand and NOT understand
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formal quiver
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formal quiver
Can anyone explain the step where they did absolute values
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ruby basin
Can anybody explain me why the result ( right side of image ) shows that the denominator has (x + 1)? I did this problem and got almost all the answers except for this denominator part, especially why it has a ( + 1 ) on it.
woeful pasture
uhh
just rewrite
1+1/x as (1+x)/x
then put the x to the nominator
so the equation is equiv to
(x^5 - 1)/(1+x)
by division
,w simplify (x^5 - 1)/(1+x)
glossy valveBOT
errant summit
maths that mental attack to mind
ruby basin
Let me digest this for a sec
wtf
let me show you my way of work and why I still dont get it
woeful pasture
Ok
ruby basin
@woeful pasture
woeful pasture
It’s the remainder
woeful pasture
Ok
So by the division algorithm
X^5-1 = quotient(1+x) + remainder
Now
Divide both sides by 1+x
ruby basin
Omg
DAmn bruh
You a math wizard
Thx
Ok so first
The first step to solving polynomial with negative exponents
Is to:
- simplify them first ( simplify the negative exponents)
- and then start dividing
- when there are remainders, just divide the remainders using the denominator from the left side of the equation
Is this a good logic?
Or what would be a way to explain in a paper ( when i am doing a test for example ) that I got this 2/(1+x) denominator from dividing (1+x) into it?
ruby basin
X^5-1 = quotient(1+x) + remainder
Or can I just write it this way after I got every results?
and then bam i got the final result form dividing both sides
brittle parcel
The first step to solving polynomial with negative exponents
first step is to write down domain restrictions
if you start out with x^-1 anywhere then x can't be zero even after simplifting go get rid of negative powers
for example
do you know the division theorem for numbers?
ruby basin
yes
brittle parcel
the one for polynomials works similarly
and you can present them the same way
13=4x3+1
13/4=3 r1
for full credit, say "using polynomial division" or "the division theorem" and thats what you did
brittle parcel
13=4x3+1
. and this
"i started with 14/6, by long division this is equal to 2 + 2/6"
something like that
ruby basin
if you start out with x^-1 anywhere then x can't be zero even after simplifting ...
Do you mean by this sentence that even after I simplified the negative exponents, it still can't be 0?
brittle parcel
Do you mean by this sentence that even after I simplified the negative exponents...
correct.
unless context suggests otherwise, you assume the domain of a function is implied by the original way the author created it
ruby basin
if you start out with x^-1 anywhere then x can't be zero even after simplifting ...
and by "anywhere" you mean both the denominator and nominator?
brittle parcel
the reason being that we don't want "simplifying" to fundamentally change the function, which changing the domain would be changing
brittle parcel
x^-1/x^-1 simplifies to
1 (for x ≠ 0)
make sense? unless context tells you otherwise, like somewhere in the problem you see that x>0 in all contexts anyway
ruby basin
Yep
brittle parcel
also
there is a super fast shortcut for polynomial long division for when you're dividing by a simple (x-c), but
ruby basin
13/4=3+1/4 is the same as this ofc
So the division theorem here ( that the remainder is actually 1/4 here) works aswell for polynomials
brittle parcel
it's not intuitive why it works at first glance
you can sesrch Google for synthetic division if youre interested
ruby basin
there is a super fast shortcut for polynomial long division for when you're divi...
Only with (x-c)?
brittle parcel
or x+c, which is just x-(-c)
brittle parcel
So the division theorem here ( that the remainder is actually 1/4 here) works as...
yes!
ruby basin
So the division theorem here ( that the remainder is actually 1/4 here) works as...
and that is why the answer for my remainder was 2/(1+x)
Now I get it
Thank you man
brittle parcel
np. synthetic division is completely optional
just a shortcut for a common case
khan academy has it
ruby basin
.close
frigid estuary
tangent line :(
full forumBOT
tangent line :(
frigid estuary
foggy vapor
This is sus
foggy vapor
What am i doing wrong
,align
y -\8{\4{\pi -4}4} &= -\8{x-\4\pi4} \[1ex]
y&= -\8{x-\4\pi4} +\4{\pi-4}4
glossy valveBOT
cedar flint
The answer is 792 right? Can someone please check
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The answer is 792 right? Can someone please check
cedar flint
To get the answer, I just did 12 choose 10 * 12 choose 11 * 12 choose 12
which is equal to 792
tight quest
The answer is 792 right? Can someone please check
No
onyx glen
To get the answer, I just did 12 choose 10 * 12 choose 11 * 12 choose 12
why multiply
cedar flint
they're independent ?
onyx glen
the number of heads was 10 or 11 or 12
cedar flint
so add?
onyx glen
yes you add
cedar flint
bruh i cant tell the difference
onyx glen
when you say "they are independent" you're making it unclear for yourself who on earth "they" are\
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thick hedge
I was thinking More than 1+ dim range(T) eigenvalues implies more than 1+ dim range(T) eigenvectors, which would mean the dimension of the range is more than dim ( range T)) which is a contradiction
thick hedge
But why is it 1+dim Range T and not range(T)
onyx glen
what eigenvalue does not "contribute" to the range
thick hedge
what eigenvalue does not "contribute" to the range
the 0 eigenvalue
onyx glen
the 0 eigenvalue
well there's your answer
gritty rose
not S_x
gritty flax
Look closer
manic bramble
First, the image is S_y, correct?
Second,
I meant to say S_x is one possible preimage of f.
stiff musk
well it can't be the preimage of S_y
fast peak
S_x equals none of the preimages of f
stiff musk
and if S_y is the whole image, then you know what the preimage of S_y has to be
fast peak
lets say the dots in X are 1,2,3 from top to bottom
and in Y are a and b
the preimage of {a} is {1,2}
the preimage of {b} is {3}
the preimage of {a,b} is {1,2,3}
manic bramble
I shouldn't have asked 2 questions at the same time. I want to get this one correct first:
Is it correct to say thatS_y is the image of f?
fast peak
yes
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storm ridge
Need some help with this grade 12 thinking question. I can of course graph it, but I’m really struggling to find an equation that follows all three of the rules.
nimble crane
<@&268886789983436800>
nimble crane
Need some help with this grade 12 thinking question. I can of course graph it, b...
have you tried anything yet? 
how about starting with the first condition? 
indigo ore
Need some help with this grade 12 thinking question. I can of course graph it, b...
Think about how the multiplicity of roots affects the local behaviour of the polynomial at that root
storm ridge
I’ve made the equation f(x) = -x(x+2)(x-4)^2 which satisfies two of the rules but no matter what I do, I can’t get the third rule to work
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storm ridge
<@&286206848099549185>
marsh oar
Use tge wavy curve analogy @storm ridge ristianThe12
analog sand
Why are we helping him
marsh oar
The*
analog sand
Isnt this 10 percent of the grade
analog sand
It says so on the paper
storm ridge
It’s open book the teacher said we can basically do whatever we want for it
queen crystal
Need some help with this grade 12 thinking question. I can of course graph it, b...
Wow, now we know your full name 😄
I’m so bad at maths
storm ridge
It’s in my username lol
sweet palm
guys i gave an extremely difficult questions of Sequence and Series. Q) If (a+1),(b+1),(c+1),(d+1) is in GP then find the value of (b+c+2)^2 = _____ options:- (a) (a+b+2)(b+c+2) (b)(a+b+1)(c+d+1) (c) (b+c+2)(c+d+2)
indigo ore
Use tge wavy curve analogy <@590631146283466763> ristianThe12
I think using that every condition is fulfilled except the decreasing part
4 to infinity is fulfilled but the other is not
storm ridge
Yeah I just can’t do (-1,1)
The only thing I can think of is messing around with the exponents until I have a point of inflection at -1 and 1 but I haven’t been able to do it
This is what I have
indigo ore
I don’t think changing the value of a will effect the x coordinate of the turning points
storm ridge
Yeah it doesn’t
I’m trying to mess with rational powers with an odd denominator but I think there’s an actual algebraic way for me to find this
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storm ridge
Need some help with this grade 12 thinking question. I can of course graph it, b...
<@&286206848099549185>
Sorry if I’m pinging too much
young night
Need some help with this grade 12 thinking question. I can of course graph it, b...
@storm ridge
storm ridge
Thanks but that doesn’t follow rule 3
sharp lagoon
!nosols
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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.
storm ridge
It also has to decrease from -1 to 1
young night
It also has to decrease from -1 to 1
it does
storm ridge
The function increases from -2 to -0.8
Also the only intervals in which the function can decrease are (-1,1) and (4,infinity) no other parts are allowed to be decreasing
glossy valveBOT
Shoshir
storm ridge
Here’s what I have I’m trying to solve for an exponent that would give an inflection at -1 and 1
My f’ might be wrong I’ve barely done any calculus
I dont think changing that value of A changes much for the inflections
glossy valveBOT
Shoshir
raw quartz
$3Ax^2-4Ax-8A<0$ when $x\in\left(-1,1\right)\bigcup\left(4,\infty\right) $
storm ridge
What does this mean
Is this question even possible?
My teacher has accidentally made multiple impossible questions before
I’m going to close this channel because there’s probably a mistake in the question that’s making it impossible
I’ll ask my teacher on Monday
.close
indigo frost
[
v_n = \int_0^1 \frac{x^{n+1}}{x+1} \dd{x} , n \in \mathbb{N}
]
prove that
[
v_n = (-1)^n(1-\frac{1}{2}+\frac{1}{3}-\cdot \cdot \cdot + \frac{(-1)^n}{n+1}-\ln(2))
]
glossy valveBOT
<rajel />
indigo frost
tried ibp , but doesnt work
vast fossil
||Sub for x+1||
indigo frost
||Sub for x+1||
,, v_n = \int_0^1 \frac{(u-1)^{n+1}}{u} \dd{u}
glossy valveBOT
<rajel />
vast fossil
And now expand
indigo frost
how would i
vast fossil
But doesn't simplify as nicely as I expected 
vast fossil
how would i
Then you just get a sum of integrals of powers of u
indigo frost
didnt understand what you mean by expand
vast fossil
(-1)^n in the front and -ln2 appear, but the rest seems a bit harder to simplify
vast fossil
didnt understand what you mean by expand
Apply binomial theorem
indigo frost
ah
vast fossil
But first let me check if this does actually work
indigo frost
btw they gave us a question which seems to be a hint
indigo frost
$\frac{x^2}{x+1}=ax+b+\frac{c}{x+1}$
glossy valveBOT
<rajel />
indigo frost
Try simplifying v_{n+1} + v_n
that would just give an integral ...
vast fossil
that would just give an integral ...
Which is way simpler, see for yourself
indigo frost
,, v_{n+1} + v_n = \int_0^1 (u-1)^{n+1} \dd{u}
glossy valveBOT
<rajel />
vast fossil
Btw the bounds of u should be from 1 to 2
vast fossil
Anyway, now you got v_{n+1} + v_n = 1/(n+2)
Yeah feel free to get back to x, the u-sub was not necessary at all
vast fossil
Anyway, now you got v_{n+1} + v_n = 1/(n+2)
The result follows from evaluating v_0 and this recursive formula
indigo frost
The result follows from evaluating v_0 and this recursive formula
but we should -vn insetad of +vn to get what you said
vast fossil
I don't think I understand what you mean
indigo frost
I don't think I understand what you mean
\begin{align}
&\text{ if we have } v_{n+1}-v_n: \
&v_1 -v_0=..................... \
&v_2 -v_1=......................\
&..................... \
&..................... \
&v_n -v_{n-1} =
\end{align}
glossy valveBOT
<rajel />
vast fossil
If you apply the formula a bunch of times, you will notice a pattern, the pattern being that v_{n+1} becomes an alternating harmonic sum (plus the initial term v_0) which can be shown via induction
vast fossil
oh you want induction
Yup
indigo frost
**<rajel />**
also this method seems to be possible
same method we got + we can get -
indigo frost
Try simplifying v_{n+1} + v_n
refering to this
vast fossil
Hmm, maybe if we wrote
\begin{align*}
v_{n+1} + v_n &= \dots \
-(v_n + v_{n-1}) &= \dots \
v_{n-1} + v_{n-2} &= \dots \
\dots &= \dots \
(-1)^n(v_1 + v_0) &= \dots
\end{align*}
And then added up, we would get a similarly elegant solution
glossy valveBOT
A Loner Bean
indigo frost
,, v_{n+1}-v_n = \int (u-2)^{n+1} \dd{x}
glossy valveBOT
<rajel />
vast fossil
Oh, you meant that
vast fossil
The integrand should be x^{n+1}(x - 1)/(x + 1) then though, no?
indigo frost
,w integrate (x-1)^{n+1}
glossy valveBOT
indigo frost
The integrand should be x^{n+1}(x - 1)/(x + 1) then though, no?
how did you get this ?
vast fossil
[ v_{n+1} - v_n = \int_0^1\frac{x^{n+2}}{x+1}\dd x - \int_0^1\frac{x^{n+1}}{x+1}\dd x = \int_0^1\frac{x^{n+2}-x^{n+1}}{x+1}\dd x = \int_0^1\frac{x^{n+1}(x-1)}{x+1}\dd x ]
glossy valveBOT
A Loner Bean
swift bridge
Not 100% sure if this is math related, but i was a bit confused with the proof regarding this edit distance problem:
swift bridge
Specifically these cases
What I don’t really understand is why for the case of inserting a character we have D[i, j-1] +1 rather than D[i+1, j] + 1
Since if we insert a character to string x, then we would have a string of length i+1 for x and not i
sudden cove
So i and j are the lengths of the strings?
And D is a function that returns the minimum number of operations?
swift bridge
So i and j are the lengths of the strings?
I believe so yes. The source I had didn’t really specify, but I think that’s what they mean
swift bridge
And D is a function that returns the minimum number of operations?
Yes
swift bridge
Wait this is recursive?
I think so. Also combined with dp (dynamic programming)
sudden cove
dp?
sudden cove
Ah okay
Are we assuming that i > j?
I also don't see how the min is coming in tbh
What is M?
What are x and y?
swift bridge
Yeah idk it’s Alr I’ll try to think a bit more
sudden cove
Okay I think I see how this works
@swift bridge
I'm going to call the strings I and J, and their lengths i and j, since x and y aren't the most helpful.
If i > j, then we need to delete a character from I and J stays the same, so we're left with D[i - 1, j]. Then we add 1 for that operation
If i < j, then we need to insert a character into I, so we have i characters left to process (since the new character is definitely fine), and j - 1 characters left to process in J, since we've processed 1 character. Then we add 1 for that operation
And for the last one, we've processed 1 character from both, and then we add 1 if they're not the same (for replacement) or 0 if they are (since we don't need to do anything)
Does that make sense?
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glad heart
Okay um so I don't get the logic behind these problems?
full forumBOT
Okay um so I don't get the logic behind these problems?
glad heart
I kinda asked for some help so I was able to solve them but I don't understand the nuts and bolts?
I guess you can break each question into two parts
a log or exponential
then you have a variable that is greater or lesser than certain numbers
How does each part influence the behavior of the graph???
analog sand
Well firstly
You should check the domain of the functions
For log x, x>0 always
So black and green will be the answers for the log part
Then you should check whether the graph is increasing or decreasing
Let us take k(x) for an example
Do you think it is inc or dec
glad heart
Um im not sure increasing?
For log x, x>0 always
What is greater than 0 ? That would be right of the y axis?
analog sand
Um im not sure increasing?
How did you deduce that
glad heart
I just took a guess, it's similar to the log x
analog sand
<@&268886789983436800>
glad heart
I didn't
glossy valveBOT
Executor
analog sand
Lol i need to learn latex
glad heart
Soemthing like \frac {1}{\1/k (x)}$
analog sand
Soemthing like \frac {1}{\1/k (x)}$
Wrong
See that the base changes when the log goes from numerator to denominator
i.e., b would become 1/b
Do you see that?
glad heart
Im not sure what you mean x would be in this?
analog sand
Im not sure what you mean x would be in this?
x would remain x
Just the base would change
Actually if you can't wrap your head around this, you dont need to do it its fine
glad heart
I'm just not sure what you mean
analog sand
**Executor**
This property isn't clear?
glad heart
"Ah well apply it now on k(x) and see what you can think of"
Okay what is a
and what is b
analog sand
That was just a general statement
a is the base, b is the argument
See if you can find what the base and argument are in k(x)
glad heart
Like one minute you're saying t oapply the formula the next you're saying not to
Say to try but say nvm
Like idk what you want
analog sand
Like one minute you're saying t oapply the formula the next you're saying not to
I said not to since you couldnt understand so we should try another method
analog sand
Like idk what you want
Have you understood the formula?
glad heart
If I had an example I could understand
analog sand
Ok
analog sand
Oh i made a typo lol
glad heart
Yeah idk lol 1/(1/k(x) except there is no log
analog sand
Yeah i just wrote the identity wrong so scratch that
See a simpler method is just
See that the base b in k(x) is between 0 and 1
That means that the higher the power of b, the lesser the resulting number will be
Can you understand that @glad heart ?
glad heart
Okay I think so
analog sand
Okay so
Do you see that when x is small (ie, closer to origin), log base b of x would be large?
Since b^some big number= a small number (x)
glad heart
Okay I think so
analog sand
Ok so you see that the curve is decreasing
Hence the answer between black and green curve is ____?
@glad heart
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deft granite
"Let D stand for the supposed data set (that is D = {D₁,..,Dₙ}). Whenever Pr(Hₙ | k) < 0.5 (where k stands for background knowledge), then generate some proposition (Pₙ) and introduce it to the conjunctive hypothesis (H = {H₁,..,Hₙ}). " im trying to build a function that does something like this
anyone know more about function stuff to help with this
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"Let D stand for the supposed data set (that is D = {D₁,..,Dₙ}). Whenever Pr(Hₙ ...
@deft granite Has your question been resolved?
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@deft granite Has your question been resolved?
deft granite
!help
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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
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@deft granite Has your question been resolved?
gritty sky
Hello 👋 over here for question 1) I solved it in the next picture and I’m wondering if my “dissecting” of the absolute value function is correct?
gritty sky
@ me when someone answers thanks !
vast fossil
Hello 👋 over here for question 1) I solved it in the next picture and I’m wonde...
Yeah, it’s correct
gritty sky
But over here the interval x>2 and the interval -2>x<=0 are the same is there a way to combine them or should they stay this way?
vast fossil
But over here the interval x>2 and the interval -2>x<=0 are the same is there a ...
You could write
[ f(x) = \begin{cases} \phantom{-}\frac{2x}{x^2-4}, &x > 2 \text{ or } -2 < x \le 0 \ -\frac{2x}{x^2-4}, &\text{otherwise} \end{cases} ]
glossy valveBOT
A Loner Bean
deft granite
bro
deft granite
"Let D stand for the supposed data set (that is D = {D₁,..,Dₙ}). Whenever Pr(Hₙ | k) < 0.5 (where k stands for background knowledge), then generate some proposition (Pₙ) and introduce it to the conjunctive hypothesis (H = {H₁,..,Hₙ}). " im trying to build a function that does something like this
anyone know more about function stuff to help with this
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@deft granite Has your question been resolved?
wooden holly
Hey there, I'd appreciate some help with this one. Unfortunately, I have no idea what to do with it, total blackness in my head.
green merlin
Hey there, I'd appreciate some help with this one. Unfortunately, I have no idea...
Are you aware of euler totient formula?
wooden holly
No, I will look into it. Thank you for the direction! 🙂
modern chasm
green merlin
Indeed
wooden holly
Oh.. another thing I've never heard of! Thank you! This question is really out of my league. 😅
green merlin
Oh.. another thing I've never heard of! Thank you! This question is really out o...
np
tardy sinew
Oh.. another thing I've never heard of! Thank you! This question is really out o...
also try different values of n and see it yourself
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true peak
can someone help explain this vectors question
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can someone help explain this vectors question
devout nova
can someone help explain this vectors question
unit vectors have a magnitude of 1
devout nova
can someone help explain this vectors question
in a 2d space, if 2 vectors are parallel then their slopes are equal
just like linear equations
plain fox
in a 2d space, if 2 vectors are parallel then their slopes are equal
srry
ig u could understand it like this , do u know what head to tail rule is?
true peak
ig u could understand it like this , do u know what head to tail rule is?
not that name but i think i know what ur talking about
plain fox
wait , is ur answer correct?
plain fox
ah ok ok , ur asking what to do next , my bad
true peak
ye
plain fox
ye
ok might have to ping the helpers role , i forgot this vector stuff
but yeah , use the fact that unit vectors have magnitude of 1 , (find magnitude of ur unit vector)
devout nova
srry
it's one way to do it
but I have found another way that does not need it
just as usual how you determine the unit vector of a vector
you divide the vector by it's magnitude to obtain the unit vector
same thing here
true peak
right i know
so I said that the magitude of OB was that thing under the root
<@&286206848099549185> could you please help me please
devout nova
so I said that the magitude of OB was that thing under the root
not necessarily
you first calculate OB
then calculate it's magnitude in terms of b
analog sand
can someone help explain this vectors question
Whats your difficulty
Take A as origin everything will be great
true peak
I don't really understand what to do next after those stes
devout nova
then calculate it's magnitude in terms of b
and you're already given the unit vector of OB
analog sand
Take A as origin everything will be great
Then O and B will resolve to points
true peak
Ok ill try with starting with A
analog sand
i added ab to oa
That would just give you the resultant vector of ab and oa
Not ob
devout nova
and then i got (b+1)i + bj + 3i
group the i components together
analog sand
That would just give you the resultant vector of ab and oa
Oh nm
Its oa not ao
So this is right
true peak
i did in the next line
devout nova
i did in the next line
oh
ok now divide OB by it's magnitude to obtain its unit vector
