#help-19
signal oar
hasty dome
They might.
One way you can do it is to fill in b.
Like if you think that b = -3/4, fill in -3/4 for b and see if that matches.
hasty dome
a(x - b)^2 + c
a(x - (-3/4))^2 + c
a(x + 3/4)^2 c.
thin anvil
I really appreciate the help
thin anvil
It's a quick way to check everything out.
Gotcha
Thanks @signal oar too
Gonna close
.close
gaunt pike
i dont understand why a⋅(1,1,1) +b⋅(1,0,1) +c⋅(−1,1,−1) = (2,0,2) isnt possible
hexed girder
And these are just points you are multiplying?
gaunt pike
yeah
hexed girder
I think this would be a system of equations problem
So, first, multiply each constant into every point
And then set up three equations
One for each direction
gaunt pike
yeah i got like, a + b - c = x, a + c = y and a + b - c = z
which means that x = z? or?
hexed girder
If it is impossible, you should get something like a + b + c = 2 in one direction, and a + b + c = 0 in another direction
hexed girder
which means that x = z? or?
Sub in the 2,0,2 into the x,y,z
gaunt pike
yeah
when im solving the system, i get that 2a + b = 2 and c = -a
oh wait
yeah nvm i got it
because c had to be equal to a, right? and not -a
hexed girder
Well it doesn’t necessarily have to be equal from what I can tell
c does have to be equal to -a
gaunt pike
yeah
brittle beacon
i dont understand why a⋅(1,1,1) +b⋅(1,0,1) +c⋅(−1,1,−1) = (2,0,2) isnt possible
Also isn’t possible? Did you write this correctly - there is an obvious choice that does make it possible!
gaunt pike
yes i wrote it correctly
pearl patio
a = 1, b = 0, and c = -1
brittle beacon
(Or a, c both being zero, and b being 2)
gaunt pike
if i solve the system i get that 2a + b = 2, c = -a and a+b-c = 2
gaunt pike
Who told you it was not possible?
chat gpt 💀
hexed girder
🤣
hexed girder
Don’t use chatgpt for maths lmfao
hexed girder
It can’t even get basic mathematical definitions correct half the time
hexed girder
then what should i used besides my brain
Your hands
gaunt pike
i am having a huge brainfart
jolly halo
then what should i used besides my brain
your teacher , trustable source ( stack exchange)
gaunt pike
how the hell am i supposed to solve this system
hexed girder
Your hands
Specifically the one you use to write with
gaunt pike
if instead of 2,0,2 i have x,y,z
pearl patio
i mean they aren't unique solutions
i think they are free parameters
a + b - c = 2
a + c = 0
a = -c
-c + b - c = 2
b = 2 + 2c
so for any b = 2+2c where you select a = -c then you'd find a solution
hexed girder
So would you just say that there are infinite solutions?
Or that it is arbitrary within these parameters?
late dust
i dont understand why a⋅(1,1,1) +b⋅(1,0,1) +c⋅(−1,1,−1) = (2,0,2) isnt possible
Why do you have x,y,z if this is your question
brittle beacon
if instead of 2,0,2 i have x,y,z
@late dust
gaunt pike
because thats for 1 one case, the og question has multiple choices
late dust
Ah
gaunt pike
so i thought that it would take a lot of time to do each case
and there must be a way to do a general case ig
i just dont know how
late dust
i have no idea what to do with this information
I mean yeah, you got the right idea here
pearl patio
So would you just say that there are infinite solutions?
yeah i guess since they're free variables then you have that there are infinitely many solutions to the system
brittle beacon
like, what happened with a b c
Basically the first column tells you about a, the second tells you about b, and the third about c
E.g. when you reduce, you get that a + b + c = x, -b + 2c = y - x and 0 = z - x
pearl patio
i have no idea what to do with this information
i guess it says exactly what we deduced, no?
you have that 0 = z- x so the system is inconsistsent for z \neq x
late dust
It basically means you cannot have a(1,1,1)+b(1,0,1)+c(−1,1,−1) = (x,y,z) if x =/= z
But y can be anything
So you can always find a solution to a(1,1,1)+b(1,0,1)+c(−1,1,−1) = (x,y,x)
(note the x in the last spot)
pearl patio
gaunt pike
we get that a = -c, b = 2 + 2c and c = c, so we can give any value to c and we get the linear combination
right?
gaunt pike
yeah yeah
im having so many brainfarts today wth
thank you all!
this wont be the last time i ask for help with "basic" things lmao
.close
odd edgeBOT
pearl patio
i ask help with basic algebra lol and addition
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mystic saffron
how do I find the height of a regular 7-sided polygon using square roots instead of trig?
south plume
likely to not be possible but i can't say for certain, a heptagon is not constructable
are you only given the sidelength?
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
likely to not be possible but i can't say for certain, a heptagon is not constru...
So you can't do it using square roots like you can with triangles or pentagons?
south plume
nope
south plume
google constructible polygons
mystic saffron
This wasn't for an assignment or anything, wanted to see if there was a relationship between the ratio between an inner circle and circles arranged tangentially around it
Triangle is 2/root 3 -1 and square is root 2 - 1
south plume
google constructible polygons
you can find the height using square roots for all of those
odd edgeBOT
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tame bronze
How would I write a quadratic equation like this in standard form?
torpid owl
just expand and distribute
sullen ferry
- foil (x-4)^2
tame bronze
then multiply that by -2?
tame bronze
Ok
clever fjordBOT
🫎 A Certain User(Moosey) 🫎
odd edgeBOT
tame bronze
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✅
tulip mica
do you know how to graph y=-2(x-4)^2+3?
tame bronze
no
tulip mica
do you know derivatives
sullen ferry
choose some values, lets say -4,-3,-2,1,0,1,2,3,4
plug those in for x
you get a y value.
that gets you where y=-2(x-4)^2+3
now you want where the y value is less then that expression
so it'll be some shaded region
harsh hedge
The values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is $10$ when $x$ is $2$, then find $x$ when $y$ is $4000$.
harsh hedge
he values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is $10$ when $x$ is $2$, then find $x$ when $y$ is $4000$.
clever fjordBOT
☆☽Darkkeeper☽☆
tulip mica
x2 and y vary inversely. let the constant between them be k. then x2=k/y
harsh hedge
ok
tulip mica
find k and then plug in the values of y and k
harsh hedge
thx I'll try that
k=40
40/4000= 1/100
x=1/10?
<@&286206848099549185>
Is it right?
nvm
it's right
.close
odd edgeBOT
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cursive raven
can someone check if my solution is right
odd edgeBOT
can someone check if my solution is right
torpid owl
8 doesn't look right
odd edgeBOT
cursive raven
idk if the - coefficient is supposed to be factored out before applying it to the coords
prretty sure this is wrong too
.reopen
odd edgeBOT
✅
white grotto
looks right to me
eternal flower
how would one solve an equation with multiple logarithms?
mystic saffron
,rotate
clever fjordBOT
toxic monolith
$\text{Hint: }\\\log_{a}x+\log_{a}y=\log_{a}\left( xy \right)\text{ where }x,y,a>0,a\neq 1$
clever fjordBOT
Joanna Angel
mortal hinge
odd edgeBOT
mystic saffron
$2\pi = \frac{10\pi}{5}$
clever fjordBOT
maxim
mortal hinge
?
mystic saffron
do you know what a coterminal angle is
mystic saffron
that's a bit vague
a coterminal angle to $\frac{34\pi}{5}$ in $[0, 2\pi]$ would be an angle such that $k + n2\pi = \frac{34\pi}{5}$ where $k \in [0, 2\pi]$ and where $n$ is a positive integer
clever fjordBOT
maxim
mystic saffron
you want to find $k$
clever fjordBOT
maxim
mystic saffron
in other words $k = \frac{34\pi}{5} - 2\pi n$, so what is $n$? how many times do we want to subtract $2\pi$ from $\frac{34\pi}{5}$?
clever fjordBOT
maxim
mystic saffron
yes
mystic saffron
ok so how do i do that
you read what I said
mortal hinge
can i js convert everything to angles or no
mystic saffron
subtract again
mortal hinge
14pi / 5 ?
mystic saffron
is that between 0 and 2pi?
mortal hinge
so 4pi/5
mystic saffron
bingo
rain tundra
How do I solve this without explicitly calculating each term? Is it even possible?
supple mural
Looks like roots of unity filter
Yeah it’s definitely ROUF
Look it up on Google
It’s hard to explain here
rain tundra
Alright
supple mural
(And I forgot how to do it lol)
odd edgeBOT
@rain tundra Has your question been resolved?
rain tundra
It's a_0 + a_3 + a_6 + a_9 + ... + a_54
rain tundra
(3)^27 = a_0 + a_1 + a_2 + ... + a_54
sullen ferry
yes
rain tundra
Yeah, it's (3)^26 given the options
sullen ferry
???
OH
WAIT
the indices
ok, try -1 and 1
i see
why roots of unity would be important here
rain tundra
So I need to put x = root of unity?
sullen ferry
yes, specifically a cube root of unity
there are 3. Two of them are conjugates of each other
sullen ferry
Yeah it’s definitely ROUF
rain tundra
hmm, solution is (3)^26 acc to this
sullen ferry
,calc e^(2pi i/3)
clever fjordBOT
Result:
-0.5 + 0.86602540378444i
,calc -.5+(sqrt(3)/2)i
clever fjordBOT
Result:
-0.5 + 0.86602540378444i
so omega=e^(2pi i/3)
rain tundra
Yeah
sullen ferry
so our formula should be 1/3(F(1)+F(omega)+F(omega^2))
rain tundra
(1+w+w^2) = F(w) = F(w^2) = 0
rain tundra
This particular could be solved without Root of Unity Filter as well as 3^27 equals sum of all coefficients and if we remove some terms from it, then the answer must be less than 3^27 which is 3^26 only among the options, so yeah even x = 1 works for this particular problem.
sullen ferry
yeah
supple mural
hmm, solution is (3)^26 acc to this
Yep, you should get 3^27/3
sullen ferry
and you know none of the coefficients a0 through a54 will be negative
so by process of elimination you could say 3^26 is the only possible answer
rain tundra
Thanks a lot!!
I guess this method will be useful for many such problems where the terms are periodic
.close
odd edgeBOT
mystic saffron
.close
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simple inlet
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simple inlet
I'm getting stuck on how to solve for v & t using the two equations on the right
also I meant to put sin25 and cos25 idk why i wrote 30
odd edgeBOT
@simple inlet Has your question been resolved?
simple inlet
<@&286206848099549185>
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@simple inlet Has your question been resolved?
regal chasm
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@simple inlet Has your question been resolved?
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last forge
Is there a way to graph complex equations like we graph x + 1/x is there a way to visualise the same for something s z + 1/z
desert marlin
what is z+1/z ?
a function of z?
and then what is z? a function of a complex variable, z=a+bi ?
odd edgeBOT
@last forge Has your question been resolved?
edgy bay
Consider a positive continuous function $g$ satisfying $\int_0^\infty \frac{dr}{g(r)}=\infty$. Suppose $x(t)$ is a differentiable function for which $|x(t)|\to\infty$ as $t\to T$ for some $T>0$. Let $x_0=x(0)$. Then apparently $$\int_0^t \frac{x’(t)}{g(|x(t)|)} dt=\int_{x_0}^{x(t)} \frac{dr}{g(|r|)}\to\pm\infty \quad t\to T.$$
I do not understand why the integral tends to positive/negative infinity.
I’d be grateful if someone could explain this in more detail.
clever fjordBOT
Philip
edgy bay
Whether or not x(0)=x_0 is positive or negative is not specified.
low locust
well the interval could effectively either be from 0 to infty or from 0 to -infty
you dont know whether x(t) goes to infty or -infty
low locust
so, you could have a few options
for example 0<x0<infty
the integral from 0 to infty diverges. the integral from 0 to x0 has to converge, cause its a finite interval and g is continuous. so also the interval from x0 to infty diverges
edgy bay
Right
low locust
and then similarly for the other examples. doesnt matter on which side of zero the point x0 is
edgy bay
But suppose x(t) goes to negative infinity. How do I then evaluate the integral?
low locust
well because of the absolute value around the r, that doesn't actually matter
more explicitly, you could do a u-sub u=-r
and then the bounds switch
odd edgeBOT
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rigid gate
in functions;
2 unique x values can have a common y value BUT
1 x value cannot have 2 y values?
lusty fiber
In a function y=f(x) ?
rigid gate
atleast in a table of x and y
lusty fiber
yes, by the definition of a function
rigid gate
in functions;
2 unique x values can have a common y value BUT
1 x value cannot h...
oh wait im correct right
modest badge
Need to simplify it basically
odd edgeBOT
Need to simplify it basically
modest badge
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odd edgeBOT
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analog epoch
coudl someone help me understand how they get from (1) to (2)? Also how do they get it to be equal to 0 at the end?
brittle beacon
How are stuff defined in there? I note they reference <w, x’> + b = 0
To which point the first term is basically 0^2 and the second term is going to be the inner product of something and zero, which is zero [oop that’s x rather than x’ they have]
odd edgeBOT
@analog epoch Has your question been resolved?
analog epoch
odd edgeBOT
analog epoch
@brittle beacon , w is a vector and b is a scalar, and these two define a hyperplane, x' is another point in the hyperplane
brittle beacon
coudl someone help me understand how they get from (1) to (2)? Also how do they ...
Looks like they substituted p into here then used inner product properties then?
analog epoch
yh but im unsure about inner product properties they used, how do they get (b + <w,x>) squared?
brittle beacon
Too lazy to tex it up but as p - x’ is -[all that stuff] + (x - x’)
Then expanding the inner product gets the first term being <-[that stuff], -[that stuff]> which is the norm of [that stuff] squared
(b + <w, x>)/|w|^2 is a constant so you can pull that out, and it becomes squared, but then you also get the norm |w|^2 left
odd edgeBOT
@analog epoch Has your question been resolved?
analog epoch
ok i got to the penultimate stage <-[that stuff], -[that stuff]> <- except at (2) i end up with denominator of the first term being equal to just 1 because when i pull out the constant
analog epoch
got to here, not sure if I have done something wrong @brittle beacon , at this point how am i supposed to get zero from here (assuming the steps i did were correct, let me knwo if they weren't)
brittle beacon
This is |w|^4 btw - squaring something that was already squared
analog epoch
oh yhh realised now, didn't notice that, thanks
ok so the first term in the last step ive written should have a | w |^2 as denominator, not sure how to progress from there tho, should i just keep expanding and see what happens,
@brittle beacon i dont get any <w, x'> + b in here, for which i could equal to 0. I mean i could say <w, x> is = -b and see if that gets me anywhere. I guess my question is how do they get from their penultimate step to 0 in the last step (for their proof)
brittle beacon
<@788085606483361802> i dont get any <w, x'> + b in here, for which i could equ...
Yea was wondering that - I think that might be worth trying? If you expand the first and second term, and do that, I think it works
First numerator is b^2 + 2b<w,x> + <w, x>^2, then the second one is just a multiplication of scalars -b<w, x> - <w, x>^2
As they’re common denominator you can simplify that down to b^2 + b<w, x> as a numerator, for which setting <w, x’> = -b would do the rest away
analog epoch
ok got it finally, thanks for the help
odd edgeBOT
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wooden swan
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modern hamlet
!status
odd edgeBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
@wooden swan Has your question been resolved?
wooden swan
```
What step are you on?
1. I don't know where to begin.
2. I have begun but go...
1
1
i wanna know the steps
what i need to do to find the answer
odd edgeBOT
@wooden swan Has your question been resolved?
drowsy root
how many rabbits are there in by the end of next year?
odd edgeBOT
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honest turtle
What do you mean by "dissonance"? Desmos is a graph tool, you need other tools for calculations.
viscid flint
the square root of something is always going to be positive (or imaginary)
honest turtle
Line 5? It does say -5, right bottom corner.
brittle beacon
If you’re assuming x real, yea
Not sure if there’s a name (if I even remember!)
but it’s a result of like the fact that multiple things have the same square, so if you square stuff in a step you can introduce extra non-valid solutions
So initially you had no solutions, but squaring created one
(Similarly with some log questions, if you’re not careful, you can also introduce/lose solutions accidentally - always a good idea to double check what you find!)
As an aside, why generally exponent and log rules require stuff to be positive
Long time no see 
winged swallow
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winged swallow
hey guys
i dont have the markscheme to this
so I wanted to ask
if my answer was correct
my answer being
$$16x(4x^2 - 3)$$
clever fjordBOT
Bananas in the sky
winged swallow
have to simplify this
so my working ->
i fournd that
$$(a+b)^6 = a^6 + 6a^5b + 15a^4b^2+20a^3b^3 + 16a^2b^4+6ab^5 + b^6$$
$$(a-b)^6 = a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$$
clever fjordBOT
Bananas in the sky
clever fjordBOT
Bananas in the sky
winged swallow
and
$$a^6 =((x+1)^{1/2})^6 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$$
$$a^4 &= ((x+1)^{1/2})^4 &= (x+1)^2 &= x^2 + 2x + 1$$
$$a^2 &= ((x+1)^{1/2})^2) &= (x+1) &= x+1$$
\intertext{Similarly}
$$b^6 =((x-1)^{1/2})^6 = (x-1)^3 = x^3 - 3x^2 + 3x - 1$$
$$b^4 &= ((x-1)^{1/2})^4 &= (x-1)^2 &= x^2 - 2x + 1$$
$$b^2 &= ((x-1)^{1/2})^2) &= (x-1) &= x-1$$
😭
clever fjordBOT
Bananas in the sky
$$a^6 &=((x+1)^{1/2})^6 &= (x+1)^3 &= x^3 + 3x^2 + 3x + 1$$
$$a^4 &= ((x+1)^{1/2})^4 &= (x+1)^2 &= x^2 + 2x + 1$$
$$a^2 &= ((x+1)^{1/2})^2) &= (x+1) &= x+1$$
\intertext{Similarly}
$$b^6 &=((x-1)^{1/2})^6 &= (x-1)^3 &= x^3 - 3x^2 + 3x - 1$$
$$b^4 &= ((x-1)^{1/2})^4 &= (x-1)^2 &= x^2 - 2x + 1$$
$$b^2 &= ((x-1)^{1/2})^2) &= (x-1) &= x-1$$
```Compilation error:```! Misplaced alignment tab character &.
l.52 $$a^6 &
=((x+1)^{1/2})^6 &= (x+1)^3 &= x^3 + 3x^2 + 3x + 1$$
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.```
winged swallow
ignoring my incompetence at latex
subbing in my values in
$$2a^6 + 30a^4b^2 + 30a^2b^4 + 2b^6$$
clever fjordBOT
Bananas in the sky
winged swallow
we get
lavish falcon
$2(x+1)^{3}+30(x+1)^{2}(x-1)+30(x+1)(x-1)^{2}+2(x-1)^{3}$
clever fjordBOT
Ruby
winged swallow
$$2(x^3+3x^2+1) + 30(x^2+2x+1)(x-1) + 30(x+1)(x^2-2x+1) + 2(x^3-3x^2+3x-1)$$
clever fjordBOT
Bananas in the sky
winged swallow
**Ruby**
yea
so
$$2[(x^3+3x^2+3x+1) + 15(x^2+2x+1)(x-1) + 15(x+1)(x^2-2x+1) + (x^3-3x^2+3x+1)]$$
clever fjordBOT
Bananas in the sky
lavish falcon
the answer is right
mystic saffron
Solve using linked lists and Pointer Machine: There is a table of size n × n. M operations of the following type are performed with it: cut out a rectangular piece of the table, turn it 180 degrees,
and paste it back. Print the state of the table after all operations are performed. Working time: O(nm).
mystic saffron
Does anybody have any ideas in this? 🥺
desert marlin
This is a computer science question not really math right?
mystic saffron
But it's related to math
odd edgeBOT
@mystic saffron Has your question been resolved?
lavish kite
i don't think there's much we can do without knowing what "linked lists and Pointer Machine" is
odd edgeBOT
@mystic saffron Has your question been resolved?
odd edgeBOT
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Available help channel!
dull plover
I really can't wrap my head around the antiderivative of 2/(e^(-x)+1)
dull plover
I've asked this question before and I still don't get it. help TT
latent scaffold
Right. Let's first rewrite it : $\frac{2}{\frac{1}{e^x}+1}$.
clever fjordBOT
Azyrashacorki
latent scaffold
Does that make sense?
latent scaffold
Not sure what you mean by that.
dull plover
ok i don't get it TT
latent scaffold
At this point, you can do two things : you can either multiply the top and bottom by $e^x$ so that you lose the $\frac{1}{e^x}$ term that is hard to deal with.
OR
clever fjordBOT
Azyrashacorki
latent scaffold
Or, you can rewrite the whole denominator so that it's on a single denominator, like $\frac{1}{e^x} + 1 = \frac{1+e^x}{e^x}$.
clever fjordBOT
Azyrashacorki
latent scaffold
And then you flip it because you have 2 over that.
In both cases, you end up with $\frac{2e^x}{1+e^x}$.
clever fjordBOT
Azyrashacorki
dull plover
then it would be
u=e^x+1
dx=du/e^x, which cancels the e^x above
latent scaffold
Indeed.
dull plover
ok yeah I kinda see it now
wow ok I need to digest this info
thanks again!
.close
mystic saffron
odd edgeBOT
clever fjordBOT
mystic saffron
Do I move everything to one side
?
And remove the natural logs since I think they cancel each other out
hasty escarp
cons more terms
mystic saffron
?
glass monolith
Is that a toilet in the background?
odd edgeBOT
@mystic saffron Has your question been resolved?
odd edgeBOT
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torpid carbon
question in nonlinear optimization:
when u got an function with an zero matrix as hessian
is the function concave or convex?
odd edgeBOT
question in nonlinear optimization:
when u got an function with an zero matrix a...
@torpid carbon Has your question been resolved?
odd edgeBOT
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clever fjordBOT
naxtisy
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
odd edgeBOT
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mystic saffron
how can i get help plz
odd edgeBOT
how can i get help plz
restive delta
Just send your question
mystic saffron
restive delta
Do you have to plot them and show or....?
mystic saffron
umm idk
restive delta
Ig you do
Anyway just draw the lines first
mystic saffron
Ig you do
Anyway just draw the lines first
for the lines for what
mystic saffron
im sorry idk how to do it can you show me plz
wooden python
aren't you the same person from yesterday who told us he can't do division, and whom i redirected to khanacademy
mystic saffron
yes
wooden python
no way you caught up all the way to here in one day, is there
mystic saffron
have faith in the math genius
because I'm missing so much work my teacher gave me everything in one day
im in gr 12 doing gr 9 work
wooden python
odd edgeBOT
@green iron Has your question been resolved?
left seal
|a-b|/a-b
restive delta
What bout it
left seal
What you think
restive delta
Idk the question
left seal
Simplfy
restive delta
What have you tried
left seal
Nothing
wooden python
$\frac{|a-b|}{a} - b$ cannot really be simplified much.
clever fjordBOT
Ann
left seal
No
wooden python
then this is unsimplifiable unless you are expected to do something trivial.
!original
odd edgeBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
signal oar
$\frac{|a-b|}{a} - b$ cannot really be simplified much.
I think he meant /(a - b), written in brackets
Then we'd have two cases
left seal
Ya
signal oar
Case 1: a - b >= 0, which is equivalent to a > b.
Then |a - b| = a - b
Case 2: a - b < 0, which is equivalent to a < b.
Then |a - b| = -(a - b)
left seal
No what you doing
wooden python
!original
odd edgeBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
left seal
Case 1: a - b >= 0, which is equivalent to a > b.
Where does 0 come from
signal oar
Where does 0 come from
I'm giving two cases. a - b being positive (or 0) and a - b being negative. One of these cases is always true, right?
a - b has to fall in either of these
If it falls into a - b >= 0, then |a - b| = a - b
If it falls into a - b < 0, then |a - b| = -(a - b)
Because the absolute value is defined as that
left seal
where does equal to 0 come from
left seal
Yeah How can it be equal to 0
signal oar
I'm not saying it is equal to 0. I'm saying there are two possibilities.
a - b is positive (or 0) or a - b is negative
Do you agree with that?
left seal
Nope
signal oar
Why not? What else can it be
left seal
Its either greater than 0 or less
left seal
No
signal oar
Is it given that a is not equal to b?
signal oar
If it's given that $a \neq b$, then it can never be $0$, that's true. Then it's either positive or negative.
left seal
Right
signal oar
In the first case, when $a - b > 0$, $|a - b|$ is just equal to $a - b$. Do you agree?
signal oar
It should be a+b?
Let's try some examples then. Say we have |8 - 5|
That's |3|, right?
left seal
Since it makes everything positive
left seal
Oh i see
wooden python
Since it makes everything positive
"makes everything positive" does NOT mean "turns all minus signs in the expression into plus signs blindly"
left seal
|b-a| is b-a
left seal
"makes everything positive" does NOT mean "turns all minus signs in the expressi...
We already over it so
signal oar
|a-b| is the same?
|a - b| = |b - a|, that's true.
left seal
What if the variable was negative
left seal
|3-4|=1
signal oar
Yeah
left seal
|-3-4| = 7
signal oar
|-3-4| = 7
Yeah, |-3 - 4| = |-7| = 7
signal oar
We could. The proof is that |b - a| = |-(a - b)| = |a - b|.
signal oar
So these are our two cases
In the first case, |a - b| = a - b.
Agree?
Because a - b is positive
So we don't have to do anything, we can just leave out ||
left seal
okay
signal oar
In the second case, a - b is negative.
left seal
Okay im ready
signal oar
Ok, so for the second case, a - b is negative
left seal
So that 2 case is negative because of what?
signal oar
Less than 0
Because that's our second possibility. Our first possibility was that it's positive
It can be positive or negative
We said that if it's positive, then |a - b| = a - b
Now we want to think about what happens when a - b is negative
signal oar
What do you mean?
signal oar
We aren't really looking at your original fraction yet, we just want to think about |a - b| so that we can go back to it and then know how to simplify it
left seal
Okay but what makes it negative all of sudden
signal oar
In the first case, 1., |a - b| = a - b
Now we want to look at the second case
And the second case says a - b is negative
signal oar
It shouldnt be possible to be negative
|a - b| is never negative, that's correct. But I'm talking about a - b here.
left seal
wait hold on
signal oar
That's just what's inside of the ||
left seal
Are we talking about a-b the divider?
signal oar
We are talking about a - b, the thing inside of the | |
As I said, we don't look at your original fraction at all yet, only at |a - b|.
left seal
Okay so
A example
|7-3|= |4|=4
|3-7| = |-4|= 4
So inside the II it can either be negative or greater than 0
And never 0?
signal oar
Yeah
left seal
Ok
left seal
no its not given
signal oar
Ok, then it can be 0 too
left seal
Just assumed so
signal oar
Like for |3 - 3| = |0| = 0 or |5 - 5| = |0| = 0
signal oar
So inside the II it can either be negative or greater than 0
Ok, so let's go through our cases.
- a - b positive
- a - b zero
- a - b negative
And look at what |a - b| turns into in each case.
In the first case, |a - b| = a - b
Right?
E.g. |5 - 3| = |2| = 2 = 5 - 3
|3 - 2| = |1| = 1 = 3 - 2
left seal
That’s positive
signal oar
That’s positive
Ok, so we know that in the first case, |a - b| = a - b.
Now for our second case, we have |a - b| = |0|
Well that's just 0
So in our second case, |a - b| = 0
Now in our third case
left seal
Negative
signal oar
we have for example
|2 - 3| = |**-1| = 1
|2 - 7| = |-**5| = 5
signal oar
Negative
Now the question is, can you think of some operation that we can perform on any negative number to turn it positive?
Like to turn -1 into 1
Or -5 into 5
without using | |
left seal
-()
signal oar
Yeah
signal oar
-()
So in this case, if we have something negative inside of | |, we can use -() to get what | | would give us (because | | always gives us the "positive version" of the number)
So e.g. if we have |2 - 3|
We know 2 - 3 is negative
So in the end, we will have -(2-3)
That's 1
signal oar
Shouldnt that be 4
Which one?
left seal
Oh never mind
signal oar
But if it was inside II itd be 1
Yeah, exactly what we get by using -(a - b)
Ok so for the a - b negative case, |a - b| = -(a - b)
left seal
Right
signal oar
Right
So we have \begin{enumerate} \item $a - b > 0$. Then, $|a - b| = a - b$. \item $a - b= 0$. Then, $|a - b| = 0$. \item $a - b < 0$. Then, $|a - b| = -(a - b)$\end{enumerate}
These three possibilities
left seal
wait hold on
It cannot be = 0
Because the variables are diffrent
If it was a-a it would work
signal oar
Because the variables are diffrent
Well, I can say a = 5, b = 5.
Just because they have different names doesn't mean their values are different
Unless explicitly stated that a not equal to b.
signal oar
Ah, you mean your original fraction
left seal
Cus the numerator would be 0
signal oar
I wasn't looking at that at all. True!
left seal
Yeah
signal oar
Cus the numerator would be 0
Denominator*, yes
left seal
ah ok
signal oar
Yeah. So if we look at our original fraction, then we just have the two cases
\begin{enumerate} \item $a - b > 0$. Then, $|a - b| = a - b$. \item $a - b < 0$. Then, $|a - b| = -(a - b)$\end{enumerate}
left seal
Okay so lets see
signal oar
signal oar
Okay so lets see
So now, we can simplify the |a - b| expression in these two cases
And so simplify the fraction
left seal
right so lets use a postive or negative
left seal
No it wasnt really needed tbh
Let’s take a case where it’s negative
-1
Then it’s also positive
No
It be negative
Because the denominator is negative
left seal
no just from examples of greater or lower than 0 from our cases for the original formula
signal oar
Ok so now first let's look at $a - b > 0$. Then in $\frac{|a - b|}{a - b}$, we can simplify $|a - b|$ to $a - b$. Agree?
signal oar
We established this here
left seal
Yes but that haas to be >0
left seal
No
left seal
We can’t do this way
signal oar
We can’t do this way
Why not?
left seal
Because the expression is always positive
It will difrrientiate the values of denminator
signal oar
Because the expression is always positive
We are doing the case a - b > 0.
We know that |a - b| = a - b in that case.
signal oar
I suggest we say a number of positive and negative if >0 or <0
We are doing the case a - b is positive.
We know that |a - b| = a - b in that case.
left seal
We know that |a - b| = a - b in that case.
Uh that should not be possible
signal oar
Why not?
left seal
Because the value of our expression could be negative
signal oar
Then a - b is positive
signal oar
Because the value of our expression could be negative
True, but we are only looking at the case a - b positive right now
We will come to the a - b negative case later too!
signal oar
5-3 = 2
Yes, positive. So this is an example for the a - b positive case
Let's look at what |a - b|/(a - b) will be
left seal
2 on denmintor aswell which is posttive
signal oar
Yeah
left seal
-2
signal oar
Yeah
left seal
2
left seal
5-3 -2
left seal
So that will be -1
Correct
signal oar
Correct
So we think that for the positive case, it will be 1 and for the negative case it will be -1 based on our examples
Now let's prove it for all a, b
left seal
1 if > 0
-1 if < 0
signal oar
Yeah
And now, we can cancel (a - b)
$\frac{\cancel{(a - b)}}{\cancel{(a - b)}} = 1$.
signal oar
No
So we proved that for a - b positive, |a - b|/(a - b) will be positive too!
Now let's go to our negative case
left seal
-1
signal oar
We think it will be -1, based on our examples. Let's prove that too
$\frac{|a - b|}{a - b} = \frac{-(a - b)}{a - b}$.
left seal
It becomes negative because denominator is negative
signal oar
left seal
Why do we have -()?
signal oar
left seal
No that’s wrong
signal oar
No that’s wrong
This is for the a - b < 0 (negative) case
left seal
But if we have a case <0 isnt it detirminator which is -1
signal oar
But if we have a case <0 isnt it detirminator which is -1
Yes, that's true. But a - b itself will be negative.
So a - b in the denominator is negative
-(a - b) in the numerator is positive
Because - something negative becomes positive
signal oar
Simplifies to -1
So for the positive case, it's 1
For the negative case, it's -1
That's the answer.
frigid canopy
What's the best way to solve |2x-floor(x)|=4 (Source ISI entrance paper 2020)
frigid canopy
rather find the number of solutions
This is what I did
let {x} denote the fractional part of x
so we have |2x-(x-{x})|=4
or |x+{x}|=4
Beyond this I'm not sure what to do
pastel dew
cool
so... x=a.bcde....
{x}=0.bcde...
you want
for positive x
a.bcde+0.bcde=4
frigid canopy
a.bcde+0.bcde=4
yeah
pastel dew
iff 2*0.bcde...=4-a
if z=0.bcde...
0<=z<1
so
0<=2z<2
so 4-a is in {0,1}
if 4-a=0
then a=4
and 0.bcde=0
if 4-a=1
a=3
frigid canopy
iff 2*0.bcde...=4-a
I didn't get what you did in the first step, here, mind explaining from there, please
pastel dew
do you get this ?
pastel dew
notice that a.bcde...=a+0.bcde.. right?
frigid canopy
Yes
pastel dew
so can you conclude the red arrow?
languid loom
odd edgeBOT
languid loom
need help asap
it's a timed assignment
And I am not understanding how to do this
I just need someone to tell me what I need to do to solve it not the answer
<@&286206848099549185>
quasi sparrow
<@&286206848099549185>
!15m
odd edgeBOT
<@&286206848099549185>
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
languid loom
!15m
thanks for the help
really appreciate how helpful you're being
it's a lot of help
quasi sparrow
it's a lot of help
Don't be entitled
hushed ore
You need to treat y = f(x)
fallow wolf
it's a timed assignment
so its a test?
hushed ore
Then when you differentiate, take use the chain rule
After which you'll need the quotient rule to find the second derivative
The correct answer is there
odd edgeBOT
@languid loom Has your question been resolved?
royal echo
I'm having trouble with this formula. My teacher didn't say much about it. How do I find the right answer for X?
pearl patio
I'm having trouble with this formula. My teacher didn't say much about it. How d...
find <PJK
then <HJK is congruent to that because vertical angles
then HGJ and HFK are similar triangles by AA similarity or anything i guess
so <G is congruent to angle F which in turn is x
of coure you can find <G from HGJ because you'd have 50 + green + G = 180 -> G = 180 - 50 - green
royal echo
of coure you can find <G from HGJ because you'd have 50 + green + G = 180 -> G =...
Oh, so G would equal 70, and since it's congruent, F would equal 70 as well. K would then equal 30, and since J equals 60, then the outer area at the top would be 40. Okay, actually I'm just confusing myself now ;-;
I think I got it for the most part
I just wrote some of the information wrongly lol
pearl patio
Oh, so G would equal 70, and since it's congruent, F would equal 70 as well. K w...
G is 70 yeah so F is 70 too
royal echo
Idk, im tired, it should be 60, not 70
pearl patio
why is K 30?
pearl patio
,calc -(50 + 70 -180)
clever fjordBOT
Result:
60
Thanks nonetheless
pearl patio
u should try to write out or draw out the informations i guess so you're not conflating stuffs
royal echo
Yeah
odd edgeBOT
Available help channel!
pure flame
If I have
I -1/n+1 I < e
odd edgeBOT
If I have
I -1/n+1 I < e
pure flame
can I ignore the minus because of the absolute value bars?
mystic saffron
no and this depends on the values of n
pure flame
man I was hoping for an easy answer im gonna lose it
brittle beacon
[if the insides were -(1/n + 1) instead, you could, but it isn’t]
pure flame
wait so I can just swoosh the negative sign and the absolute value bars
and keep going yye?
brittle beacon
-1/n + 1 is usually to be interpreted as -(1/n) + 1
brittle beacon
wait so I can just swoosh the negative sign and the absolute value bars
Yep you can 
brittle beacon
Yep, that’s looking good to me 
odd edgeBOT
odd edgeBOT
Available help channel!
open knot
if f is continuous on some closed interval [a,b], does that mean inf{f} = min{f} on that interval
brittle beacon
Something something, a theorem that you attain your minimum/maximum when you're defined on a closed bounded interval
ember oak
if f is continuous on some closed interval [a,b], does that mean inf{f} = min{f}...
This is a property of compactness too I believe. @brittle beacon am I right?
open knot
oh k, will look that up briefly
@brittle beacon & sowy for multiple occupations, my main question will probably maintain its place for a bit 🪼 ˢʰᵃᵐᵉ ᵒⁿ ᵐᵉ
brittle beacon
This is a property of compactness too I believe. <@788085606483361802> am I righ...
If I recall right, I mean compactness is equivalent to being closed and bounded 
open knot
This is a property of compactness too I believe. <@788085606483361802> am I righ...
ah right, since continuity implies boundaries and boundaries imply that minimum and maximum exist
kk :) thank thees
.close
odd edgeBOT
brittle beacon
ah right, since continuity implies boundaries and boundaries imply that minimum ...
Though be a bit careful - you need both the continuity and boundedness (though can't think of any non-artificial examples haha)
odd edgeBOT
Available help channel!
frosty stone
odd edgeBOT
