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#help-13
crimson sedge
livid monolith
cedar kilnBOT
livid monolith
What am I doing wrong?
glad kestrel
where'd you get x-2 and x+1 from
livid monolith
(X+2)^/ (x-1)^2
glad kestrel
(x+2)^2 is not (x+2)(x-2)
same with (x-1)^2
that being said, the 2 should just be brought to the front using log rules
livid monolith
that being said, the 2 should just be brought to the front using log rules
2log[(x+2)/(x-1)]
N then what?
glad kestrel
wait i input that wrong
ah i see
i was wrong before, the two should not be brought to the front
it should be used as an exponent
and anything squared will be positive
cedar kilnBOT
@livid monolith Has your question been resolved?
livid monolith
and anything squared will be positive
Ahhh, so he just wrote it in a fancy way to describe R-{-1,2}
violet flume
grab a new channel
pliant orbit
bro is mine
violet flume
no i mean
itll bugg out
if you try to post now
just post in a new one
its about to close this
pliant orbit
oh
glad kestrel
it may be sent to the shadow realm 
pliant orbit
ok
glad kestrel
huh
crimson sedge
I'm having doubt regarding divisibility of integers
So if a|b and b|c then a|c can someone show an example of it?
if 3|6 and 6|18 then 3|108 idk how 3|108 was got
crimson sedge
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
@crimson sedge Has your question been resolved?
lunar light
Hi, im good at maths, but i work very slowly, often I find that I know exactly how to do a question, yet it takes me 20mins for 6marks, when exam pace is 1mark/min
this makes it very hard to get much work done, Im retaking A-level exams in 2months, and I have to learn A LOT of content, single, further maths + physics (revising year 1, and self teaching year 2 from scratch)
I am looking for ways to become significantly faster, preferably transferable skills, so that i become naturally faster at all questions, not just the ones i practice all day
im not sure if it's writing speed or distractibility, or not knowing enough mental maths/shortcuts
i love maths and am VERY scared i wont be able to pursue it at uni due to my speed despite deep understanding
cedar kilnBOT
Hi, im good at maths, but i work very slowly, often I find that I know exactly h...
@lunar light Has your question been resolved?
celest ledge
Maybe it’s not the speed pf solving questions but the speed of reading materials that we should worry about…
cedar kilnBOT
lunar light
.reopen
cedar kilnBOT
✅
tropic oxide
it is most likely that you're spending more time than necessary on arithmetic
i don't yet have enough data (or any, really) to assert that with any confidence
but i've seen a lot of people waste time doing more arithmetic than necessary
hollow osprey
^ along with what she said, practice makes perfect. this video could offer some insight:
https://www.youtube.com/watch?v=BqWqzvnbnjU
sharp crane
Pretty sure you can use calculators on all a level papers now, though. A lot of a level papers tend to just be the same questions more or less with different numbers each year, though. So it might be a matter of figuring out which question you're being asked and then running through the standard method for it. That's typically what it comes down to.
fallen heath
Hi, im good at maths, but i work very slowly, often I find that I know exactly h...
Hi, im good at maths, but i work very slowly, often I find that I know exactly how to do a question, yet it takes me 20mins for 6marks, when exam pace is 1mark/min
this makes it very hard to get much work done, Im retaking A-level exams in 2months, and I have to learn A LOT of content, single, further maths + physics (revising year 1, and self teaching year 2 from scratch)
I am looking for ways to become significantly faster, preferably transferable skills, so that i become naturally faster at all questions, not just the ones i practice all day
im not sure if it's writing speed or distractibility, or not knowing enough mental maths/shortcuts
i love maths and am VERY scared i wont be able to pursue it at uni due to my speed despite deep understanding
Looks like you're lacking at organization skills (!?)
lunar light
Pretty sure you can use calculators on all a level papers now, though. A lot of ...
thats not my problem, i know the standard method, it just take forever to execute it
fallen heath
Do you have a priority set up, when doing a paper, or any question in general?
lunar light
wdym priority setup
idk it might be my handwriting, imma email the learning support of my old school later today
lunar light
Looks like you're lacking at organization skills (!?)
yes my organisational skills are very questionable
fallen heath
._. not quite, I had a peer whose handwriting was literally chicken pecks but he'd be the second fastest in the class at calc questions
But anyways, I'd like to first confirm if your exam comes as a combined paper mixed with questions from all topics or math, physics, etc. individually on different dates?
lunar light
we have seperate papers for each module, each on different dates
modules can be pure y1, mechanics e.t.c
but the actual topics in each module varies
fallen heath
and what might be your issue then(!?) not being able to solve the paper at the optimal exam pace?
lunar light
i can't do the test quick enough, nor can i practice book exercies quick enough
fallen heath
if it's that, I'd say it's the lack of practice, and not a hard brained practice but an efficient organized one where you keep in mind the types of questions and are capable of recognising those on sight
lunar light
i don't think that would account for 1/3 speed relative to peers/ exam speed
fallen heath
I'm pretty confident that's indeed the case. Because no matter how fast your handwriting speed is, the proper steps at an exam must be neatly mentioned so lacking at that would only probably account for 1/6th the time if you've been consistently fluently walking through the paper
scenic delta
if it's that, I'd say it's the lack of practice, and not a hard brained practice...
any strats on doing this?
fallen heath
Furthermore, not being familiar with a lot of question patterns costs you time to first make you do that think out of the box process, which further delays you during an exam
crimson sedge
Hi, im good at maths, but i work very slowly, often I find that I know exactly h...
this is math question for improving math
crimson sedge
this is math help for math questions
scenic delta
fallen heath
it's no different than playing chess, at first you randomly guess appropriate moves depending on your situation and your knowledge of valid moves, and what might work
but slowly, and steadily, your knowledge of your situations increases
as you play more
and you start recognising possible traps that you'd rather avoid,
and the efficient moves become a habit
same with math tbh
steel canopy
Is there a way to improve percentage of passing linear circuit analysis II?
steel canopy
Yea I don't think think would help if 100 percent of grade are test
fallen heath
Given $x + 2 = 3$ , you wouldn't write that as $2 = 3 - x$ or $1+2/x = 3/x$ to solve for x
wraith daggerBOT
crimson sedge
and the efficient moves become a habit
Long story short u wanna say " Practice"?
fallen heath
that's what I mean by habit
steel canopy
I wish there someone to tutor linear circuit, but I can barely find anyone.
fallen heath
I see linear circuit, I think resistors in series
steel canopy
Physics is annoying, but circuit is liked another level of nightmare.
lunar light
thankyou for helping me 🙂 i will continue with the grind
steel canopy
Like you have E^jw+phase shift.
fallen heath
Given the sum of 1331 and 1332 terms of an AP
And when you're asked to find the 1332th term
the one, first idea that comes to mind is:
$$\frac{1331}{2}(2a + 1330d) = S_{1331}$$
wraith daggerBOT
fallen heath
$$\frac{1332}{2}(2a + 1331d) = S_{1332}$$
wraith daggerBOT
fallen heath
and you then write the formula for $a_{1332}$
wraith daggerBOT
fallen heath
and use the other two equations to figure the answer
but hell no, that already frickin cost you more time and brain than your peers
Just realize $$S_{1332} = S_{1331} + a_{1332}$$ and you're done
wraith daggerBOT
fallen heath
that's one step versus idk, half a page
or even more
and mind you,
this is not due to practice
or just hard braining your way through the exercise problems
Cause you might've just as well done the same exact process through all such questions in your exercise problems
This is due to being conscious during your time doing your grind
continuously observing common patterns, or points.. looking out for the potentially intended solution
or the thought process
steel canopy
When you are freshman in college you looked lively, but when you are sophomore in college you are serious. When you are junior in college you are half dead, but if you are in senior in college you looked like dead corpse. That's what I heard from graduate students that's how you determined the year in college.
waxen laurel
cedar kilnBOT
gray rose
hm
waxen laurel
It's so difficult
gray rose
ik
cedar kilnBOT
@waxen laurel Has your question been resolved?
waxen laurel
<@&286206848099549185>
cedar kilnBOT
waxen laurel
.reopen
cedar kilnBOT
✅
waxen laurel
Sorry i solved it wrongly
cedar kilnBOT
@waxen laurel Has your question been resolved?
waxen laurel
<@&286206848099549185>
waxen laurel
<@&286206848099549185>
cedar kilnBOT
@waxen laurel Has your question been resolved?
wide light
cedar kilnBOT
wide light
Part ii
Question 2
I'm having trouble working out the displacement equation
I think I'm getting it wrong
frigid canyon
send what you have done
wide light
So I got -175/6 for my c value for s2
I don't think thars right at all
I think it's supposed to be 195/16
quaint turtle
well, that's what a calculator is for 
wide light
Have I used a calculator wrong?
Which equation?
Let me show my calculator working for the 1525/48, I got a feeling that's the issue
wide light
wide light
the method's wrong
Wdym?
I'm integrating to find displacement equation
And subbing respective c values in
frigid canyon
there's no need of c i think
wide light
One second let.me send something
My teacher went through this question quickly in class a few weeks ago
These where the scribbles/notes I made on it
I don't understand them because I don't remember how I got that 195/16
Oh wait
I might know the issue
I think I worked out the c for my V2 value incorrectly?
Yes I did
It should be 195/16
Let me try again
sand ether
whats the question
cedar kilnBOT
@wide light Has your question been resolved?
wide light
@sand ether
up here
im still having issues with it
im doing it the method my teacher told me to do
but im not sure its correct
so by setting s2 = 100
i got 26.8897
this is the way my teacher told me to do it, im not sure if thats the correct answer though
however, if i do it using both equatiosn
sand ether
You're aware that $v = \int a~~ dt$?
wide light
yes
ive done all that
a = acceleration
v = speed
s = displacement
12 for corresponding equations subscript
ok back to what i was saying
if i do it using 2 equations
at 2.5 seconds, s1 = 125/6
sand ether
wide light
if i set s2 = 100 - 125/6
then solve for time
i get 27
then i need to add on 2.5 onto that
which is a different answer from the previous method
there is my issue.
sand ether
So you have $v=10t-2t^2 ~~ 0\leq t \leq 2.5$ and $v=\frac{t}{4} - \frac{t^2}{20}+k$
wraith daggerBOT
azeem321
wide light
no
sand ether
compare them
wide light
i can do parti
part ii is my issue
ive explained my exact issue above
how i get 2 different answers with 2 different methods
wide light
displacement equations here
theres a velocity turning point at t=18 i think
my teacher seemed to ignore that though
cedar kilnBOT
@wide light Has your question been resolved?
wide light
I got this
Why is it incorrect to do my other method
and also
why does the 18 second turning point not matter
cedar kilnBOT
@wide light Has your question been resolved?
cedar kilnBOT
@wide light Has your question been resolved?
cedar kilnBOT
@wide light Has your question been resolved?
cedar kilnBOT
@wide light Has your question been resolved?
sand ether
I imagine no responses because: no latex and hairy numbers.
im bad at reading handwriting hence why i cant help much
silent bobcat
I imagine you could already answer @wide light
Who was helping you all along yesterday
wide light
I imagine you could already answer <@!558061348358193163>
I can answer but
My question was why will it not work with the alternative method
silent bobcat
I was talking to azeem, not you. It was just in order to refer to you
wide light
I will write out my alternative method question again iab if it's got pushed off
Out rn
silent bobcat
👍
cedar kilnBOT
@wide light Has your question been resolved?
crimson sedge
@wide light if ur done u can close the channel with .close
cold shard
<@&268886789983436800> Are you able to manually close the channel? The torrent guy’s had it active to himself for days..
calm sierra
certainly unusual but theres no harm in keeping the channel
we have many other channels to keep up with demand
calm sierra
i do suggest to @wide light to repost the question, it may be buried under the convo
wide light
Part ii is the question but here's the thing
I can do the question by making 1 equation for displacement which is from a2 (the second acceleration equation) and then integrating it twice and taking the two constants of integration from a1
This is also the way that someone else did it that helped me and my teacher told me to do it
My question is, once you have integrated twice both equations, why do you only use the 2nd equation to find the entire time even though between 0 and 2.5 is outside the range of the equation
For example, if you was to calculate how far the runner runs in the first 2.5 seconds using s1
Then subtract that from 100 and set that equal to s2
Then add 2.5 to the value of time you get from s2, that is a different answer (and incorrect). Why?
There is also a v=0 at about 18 seconds.
This may be the reason the 2nd method don't work, but it is ignored in the first method anyway for some reason??
fallen heath
I can do the question by making 1 equation for displacement which is from a2 (th...
taking the two constants of integration from a1?
wide light
Yes
My working out for the 'corrrct' method I've posted previously
Actually the working out may be wrong
wide light
I got this
Results in this answer though which Azeem also got
fallen heath
Okay
wide light
But I find it wierd that you use the originally incorrect model for the entire question, and you ignore the turning point.
fallen heath
Then subtract that from 100 and set that equal to s2
this idea
where'd you get that?
s2 is only applicable for t > 2.5
wide light
And using both models (when you integrate both equations twice and use each one for it's respective tim interval) is incorrect
wide light
s2 is only applicable for t > 2.5
But apparently you use s2 for the ENTIRE distance
Which is what I don't get
fallen heath
because
for the ENTIRE distance that you mention
"s2" is only accountable for the distance covered after the first 2.5 seconds
say s/he ran for 7 secs
then s1 is integral from t = 0 to 2.5
and s2 is integral from 2.5 to 7
wide light
But in order to get the answe
R
Per the correct method
You do s2 = 100
And solve for t
wide light
Yes, because that's what I did when I got the answer, and that's what my teacher did
fallen heath
what is your s2?
wide light
And that got the correct answer
fallen heath
for some reason, I'm getting that he finishes the race within the first 2.5 seconds?
da heck, that's world record
a = 10 - 4t
v(t) = 10t - 2t²
x(t) = 5t² - 2t²/3
for first 2.5 seconds
@sand ether can you help me out here?
sand ether
If you can't do it, what makes u think i can 
fallen heath
You figured the answer though
okay okay, I was right oof
,w integrate (2.5 - t)/10 from t = 2.5 to x
wraith daggerBOT
fallen heath
,calc 5(2.5) - 0.3125
wraith daggerBOT
Result:
12.1875
,w integrate 12.1875+0.25t-0.05t^2 from 2.5 to x
wraith daggerBOT
fallen heath
,calc -30.9896 + 20 + 5/6
wraith daggerBOT
Result:
-10.156266666667
the thing is
when using $v(t)$ for $t > 2.5$ while integrating for $x(t)$
wraith daggerBOT
fallen heath
You use this: $$x_t - x_{2.5} = \int_{2.5}^t v(t) \dd t$$
wraith daggerBOT
fallen heath
so, your displacement curves for both s1 and s2 are connected
they look like this
the red one is for x < 2.5 and green is for x > 2.5
fallen heath
For example, if you was to calculate how far the runner runs in the first 2.5 se...
this however will not work
because, the curves, even if connected, are not the one and the same
C1(red curve) accounts for the displacement from 0 for the first 2.5 secs
C2(green curve) accounts for the displacement from 0 at any t > 2.5 secs
so, when equating x(t) = 100, the s2 works because 100m was not achieved within the interval of time for s1
consider resolving the question on your own once and you'll understand
especially the integration part
wide light
Thanks, im thinking about it i think i get it..
fallen heath
I sure hope so!
x(t) is the displacement at time "t"
x_1 (t) and x_2 (t) are different curves
x_1 (t) is the displacement curve for 0 < t < 2.5
x_2 (t) is the displacement curve for t > 2.5
dire geode
.close
hollow mica
Ok so here's my issue. I'm trying to get rid of that sigma ln xi on top because it's screwing up my equation to find the cramer rao lower bound with my estimator
Ideally it should be gone altogether somehow but I don't know if I made an error or what.
If this equation made any sense, I'd have E(1/y) part
hollow mica
cedar kilnBOT
@hollow mica Has your question been resolved?
cedar kilnBOT
@hollow mica Has your question been resolved?
dire geode
Ok so here's my issue. I'm trying to get rid of that sigma ln xi on top because ...
Where's the original question
jaunty mural
$(a+b) - (c+d)$
wraith daggerBOT
Shuri2060
jaunty mural
Can you get rid of the brackets for me on this please.
night sedge
@red epoch HI so this is the problem
cedar kilnBOT
<@!708669926398754856> HI so this is the problem
red epoch
ok
night sedge
red epoch
do you know how to start this
red epoch
yeah do that
humble moss
pretty sure there is 3 solutions
red epoch
I got 88x - 38x^2 + 4x^3 = 60
night sedge
yup same
and i moved 60 to the left and set zero as the solution
but idk what to do after that
humble moss
i think 1
red epoch
1??
glad kestrel
that's not a trinomial
night sedge
LOL IDEK
hollow mica
Where's the original question
It's fine now.
red epoch
that's not a trinomial
what is it
glad kestrel
a polynomial...?
red epoch
a polynomial...?
yeah thats right
humble moss
wait do u know how to solve it
get all terms onto one side and use quadratic formula
or you could also set each term = 60
red epoch
get all terms onto one side and use quadratic formula
this isn’t a quadratic
night sedge
this isn’t a quadratic
^^
glad kestrel
probably use the factor theorem
red epoch
Try like breaking it up into 3 parts, starting with 2x, 2x, and z.
red epoch
probably use the factor theorem
what’s that
humble moss
oh theres a cubed term
glad kestrel
p's and q's
humble moss
$4x^3-38x^2+88x-60=0$
wraith daggerBOT
Vladilena Milizé
sharp kestrel
cedar kilnBOT
glad kestrel
is that supposed to be floor
sharp kestrel
is that supposed to be floor
?
glad kestrel
$\floor{x}$
wraith daggerBOT
a disappointing son
glad kestrel
your teacher is correct
[x] just means go to the nearest integer that is less than or equal to x
sharp kestrel
so what did I do wrong
glad kestrel
-17.1 isn't an integer
jaunty mural
[x] just means go to the nearest integer that is less than or equal to x
I have doubts.
glad kestrel
jaunty mural
It can vary upon definition
glad kestrel
according to the teacher's answer i think that's what they want the student to do
jaunty mural
No idea what you mean by this.
sharp kestrel
the largest integer not bigger than x
jaunty mural
nvm misread
jaunty mural
the largest integer not bigger than x
ok
Then proceed to find what that is for -17.1
yes?
sharp kestrel
uh
its -18?
so
I have to write what I did wrong
so what do I say
do I say I didn't find the integer for -17.1
@jaunty mural
.close
leaden crater
cedar kilnBOT
leaden crater
is there a way to solve this without knowing about ramanujans inf radical
leaden crater
.close
leaden crater
cedar kilnBOT
leaden crater
is there a way to get that first line without using trial and error
the line where it says $ f(x+3)=(x+3)^{3}+2(x+3)^{2}+3(x+3)+5 $
cedar kilnBOT
@leaden crater Has your question been resolved?
cedar kilnBOT
@leaden crater Has your question been resolved?
leaden crater
<@&286206848099549185>
wraith daggerBOT
aspwil
wraith daggerBOT
aspwil
wraith daggerBOT
aspwil
crimson sedge
@leaden crater
What is g(x-3) when we plug it into the first equation
To refrase
What is $ g(x-3) $ given $g(x) = f(x+3)$
wraith daggerBOT
aspwil
crimson sedge
@leaden crater you still here?
leaden crater
Why g(x-3)
crimson sedge
**aspwil**
Plug this in and you will see
leaden crater
x^3 + 2x^2 + 3x + 5?
Makes sense but how come we do x-3
Like what leads us to that
wraith daggerBOT
aspwil
crimson sedge
Were undoing the +3 part in the definition of f(x+3)
wide hound
cedar kilnBOT
wide hound
and that will give my a value?
quartz agate
Yeah
wide hound
the answer was 7
quartz agate
I got 7
wide hound
wait u did?
quartz agate
Yeah
wide hound
do i sub x=2 into the y'(x) equations
quartz agate
Actually this question doesn't need derivative
wide hound
wait how do uk?
cuz thats the thing i get confused about
like when do I need to derivave
wide hound
but lets say on a test i was given this
when would I know that I need to derivave
or not
if that makes any sense
quartz agate
If the gradient of the linear equation is given
Then the question mentions that the linear line is the tangent of that function at a point
Then I think you can do the derivative
Use this question as an example, if the linear line is 7x-4y+21=0
And it's a tangent to y=a/x² at x=-2
Then you can calculate the derivative
humble moss
The deriverative gives you the slope at a certain spot
quartz agate
Well if you use derivative
It is used to find the gradient at that point
And you will get
$$\frac{dy}{dx}=\frac{a}{4}$$
wraith daggerBOT
skypirate
quartz agate
After you substitute the x=-2
@wide hound what will you get if you rearrange the linear equation to y=mx+c?
cedar kilnBOT
@wide hound Has your question been resolved?
wide hound
y=a/4?
wide hound
yes
quartz agate
And then substitute X=-2 into the linear equation
Then you will have two equations
Solve them simultaneously
cedar kilnBOT
@wide hound Has your question been resolved?
main current
cedar kilnBOT
tropic oxide
yes this is ok
cedar kilnBOT
@main current Has your question been resolved?
main current
yes this is ok
Ok thank you
tawny sundial
in ax^2 + bx + c, can b be greater than or equal to ac?
cedar kilnBOT
in ax^2 + bx + c, can b be greater than or equal to ac?
tawny sundial
(in a quadratic)
fallen heath
what do you think
tawny sundial
wait can it be like
(ax + 1)(x + c)
ax^2 + x + acx + c
so 1+ac
so b can be no more than 1 + ac?
nvm i think that's alright
.close
cedar kilnBOT
fallen heath
wait can it be like
(ax + 1)(x + c)
ax^2 + x + acx + c
so 1+ac
so b can be no mo...
not quite true
b can be anything
$$4x^2 + 6x + 1 = 0$$ is also a quadratic
wraith daggerBOT
cedar kilnBOT
Available help channel!
earnest storm
If I have two different functions, how would I calculate the maximal linear distance within an interval?
fair plover
say f(x) and g(x)
find x when d/dx {f(x)-g(x)}=0 to find the min/max points
test all min/max point and start/end point of interval
earnest storm
How would I insert the intervals mathematically?
Like if I want to find the maximal linear distance between 4-4.5
fair plover
if d/dx {f(x)-g(x)}=0 does not have a real solution then test f(4)-g(4) and f(4.5)-g(4.5)
whichever has greater value has a further distance
earnest storm
if d/dx {f(x)-g(x)}=0 does not have a real solution then test f(4)-g(4) and f(4....
The thing is, I wanna find the maximal linear distance between 4-4.5
So I don't know whether the value could be 4.1, 4.234, or something else
fair plover
therefore you have to do the derivative test for {f(x)-g(x)}
and try to find maximum points
fair plover
so f(x) is the first function and P3(x) is the second function?
earnest storm
Yeah, dw
fair plover
at x=3.622, it is a turning point for {f(x)-g(x)}
when the derivative test is carried out, youll find out that x=3.622 is a minimum point.
therefore to find the maximum separation you just have to test the start and end point of the interval
namely, x=3 and x=4.245 for {f(x)-g(x)}
whichever have a larger number, thats the maximum separation
earnest storm
So what would the maximum linear distance be between the functions
In the given interval [3,4.245]
fair plover
0.0398205577408 if youre only looking for the answer but not the method
earnest storm
0.0398205577408 if youre only looking for the answer but not the method
What's the method you've used?
I'm looking for the method
fair plover
-
d/dx {f(x)-g(x)}=0, x=3.622
-
carry out derivative test, x=3.622 is a minimum point
this implies the distance between f(x) and g(x) decrease when x=3 to x=3.622, then starts to increase when x=3.622 to x=4.245 -
so the max separation could only be x=3 or x=4.245
-
f(3)-g(3)=0.0278622936473, f(4.245)-g(4.245)=0.0398205577408
-
so the max separation is when x=4.245, and the separation is 0.0398205577408
earnest storm
I'll get back to you in one second
I'm trying to test it out, whether or not it checks out with the results list
Appreciate the help @fair plover 😄
fair plover
glad to be helping :)
cedar kilnBOT
@earnest storm Has your question been resolved?
crimson sedge
Can someone tell how I can calculate x? (q,m are constants) I need a mathematical way to calculate x. I know that I can write it as x * (q - q div m) but then im not finished yet, what has to be done ? I usually do that simplification (x * (q- q div m)) instead of (x * q) and then its fairly simple to think about the solution but I need a full mathematic way
soft cloak
Finding an inverse modulo m is closely related to finding solutions for Bézout's identity
i.e finding (x,y) such that xq+ym=1
Modulo m you indeed get xq≡1[m]
Bézout's theorem is mostly an existence proof though, and there aren't any specific formulas, but the Extended Euclidean Algorithm computes them very efficiently
crimson sedge
oh yeah I know of that algorithm @soft cloak , though could you explain to me a bit more detailed how I can find the inverse and why it works
soft cloak
You mean, how the algorithm works, or why and when it exists?
Oh yeah, i nearly forgot to mention, the inverse might not always exist depending on the choice of q and m
cedar kilnBOT
@crimson sedge Has your question been resolved?
crimson sedge
You mean, how the algorithm works, or why and when it exists?
I know how the algorithm works, though I wonder why the inverse exists and when it exists
soft cloak
I see
Well, the key fact is Bézout's theorem
That states that for any two integers a and b
There exist integers x and y such that ax+by=gcd(a,b)
Now, if a and b are coprime, gcd(a,b)=1
So you get x and y such that ax+by=1
If you go mod b, you get ax≡1[b]
Boom, here's an inverse of a mod b
crimson sedge
Oooh 🧐
soft cloak
And that tells you that the numbers that have inverses mod b are precisely the numbers that are coprime with b
crimson sedge
Thanks, I see now. and yeah, if e.g. q = 70 then x * q mod 5 can never be 1 for all x
in that case you wont have a solution right?
sorry forgot the mod 5
edited
soft cloak
correct
a way you can see this
Look at 70 and 5 and the values that 70x+5y can be
Since 5 divides 70 it's not hard to see that you can reach every multiple of 5
And in fact, you can only reach multiples of 5
You can see this as 70x+5y not being able to comb the integers beyond multiples of 5
4x+6y can't comb the integers beyond multiples of 2
However, 3x+7y can reach all integers
crimson sedge
yeah?
soft cloak
the gcd is often introduced as, well, the "greatest common divisor"
But there's another (equivalent) definition which really makes Bézout's theorem stick out
You can prove in group theory that the aℤ (for a∈ℤ) are the only subgroups of ℤ (i.e sets of numbers that are stable by addition and taking the opposite)
And also that for any integers a and b, aℤ+bℤ (that is, the set of ax+by for x and y integers) is also a subgroup of ℤ
Hence, for any a and b, there exists some integer d such that aℤ+bℤ = dℤ
And that d is precisely gcd(a,b)
This is entirely equivalent to the usual definition (proving that takes some slight work though)
And you can see how bézout's theorem comes from this
gcd(a,b) ∈ aℤ+bℤ, thus there exists x,y∈ℤ such that gcd(a,b)=ax+by
soft cloak
This definition of the gcd blew my mind honestly
The lcm is defined similarly
aℤ∩bℤ is also a subgroup of ℤ, so aℤ∩bℤ=mℤ and m=lcm(a,b)
it's delightfully elegant
that's all i wanted to show
marble latch
cedar kilnBOT
marble latch
I’m not getting the answer !?!!?!
queen horizon
$$r = \dfrac {1}{3} $$
marble latch
Yeah?
wraith daggerBOT
culdesac
fallen vine
How did they get the answers they got here?
cedar kilnBOT
How did they get the answers they got here?
@fallen vine Has your question been resolved?
thick oasis
How do I rigorously prove that a multi-variable function can output a natural number given the right input?
cedar kilnBOT
How do I rigorously prove that a multi-variable function can output a natural nu...
@thick oasis Has your question been resolved?
leaden snow
@thick oasis Not necessarily always possible depending on domain and range of the function
do you have a specific function?
thick oasis
<@382977855920930816> Not necessarily always possible depending on domain and ra...
Hm, fair enough ig
leaden snow
or constraints?
thick oasis
do you have a specific function?
Yep
leaden snow
Show?
thick oasis
leaden snow
a,b,c can be all any real number?
thick oasis
Yeah, but c is always a natural number
leaden snow
hmm
so the cube roots at least need to yield something rational, or even better if the terms under the radicals are perfect squares/cubes
thick oasis
so the cube roots at least need to yield something rational, or even better if t...
yep. The + p at the end helps a lot too, so stuff like 2/3 + 1/3 = 1
leaden snow
are there perfect cubes q + u and q - u?
thick oasis
are there perfect cubes q + u and q - u?
u?
leaden snow
under the third root
thick oasis
mkay
leaden snow
i just let u be the rest after q + ..
thick oasis
there are perfect cubes, at least if you consider rationals
with a = 5, b = 2 and c = 2
I've done some testing
cedar kilnBOT
@thick oasis Has your question been resolved?
sweet ingot
Heyy
fallen solar
I need help with solving this
solve? or simplify?
because there's no equation to "solve" here
sweet ingot
I meant simplify and reduction
sweet ingot
solve? or simplify?
^
fallen solar
^
then factorising the expressions in the numerator and denominator would be a good start
fallen solar
ye
sweet ingot
and what would be the denominator of it then
because i can't think of a number that doubled of divided by 17
except for 17 ofc
fallen solar
try splitting to 14 and 3
sweet ingot
i dont think i understood
fallen solar
2m^2 + 14m + 3m + 21
sweet ingot
how does this helps me exactly
fallen solar
how does this helps me exactly
uhm you could factorise the corresponding terms
by taking some things common
fallen solar
i still can't make it work
ok how do you usually factorise expressions
fallen solar
yup
sweet ingot
so now i'm left with
2m+3/2m+3 * 3
that means i can take both of 2m+3 down
correct?
so my answer that i'm left with is 3
fallen solar
right
sweet ingot
.close
sonic dust
Hey! Can u find the value of x here? with reasons possibly
cedar kilnBOT
Hey! Can u find the value of x here? with reasons possibly
sonic dust
crimson sedge
mfs aren't helping me prob not gon help you either :(
cedar kilnBOT
@sonic dust Has your question been resolved?
leaden snow
@sonic dust Angle Q & S are the same and angle P & R are the same
sonic dust
yes
i know that
Q = S = 70
P = R = 110
those 2 are scalene triangles, so how do i find x in this situation?
leaden snow
so the total interior angle sum is 360
sonic dust
yup
leaden snow
one sec lemme write this down on paper
sonic dust
360 - (70 + 70) = 220 / 2 = 110
so Angle P and R is 110 each
sonic dust
one sec lemme write this down on paper
kk
leaden snow
pretty sure you cant. consider a single triangle with one angle 70 deg. you can scale the legs such that the other angles can assume "arbitrary" values. then copy that triangle and glue it onto the original one to make a parallelogram.
so x is not really determined
sonic dust
Well i dont really understand the example you said, but I dont think i can find the value of x unless i find another value of one of the triangles if not then theres 2 unknowns and can't really solve it
sonic dust
so x is not really determined
I see
leaden snow
yes
sonic dust
maybe the book has a problem
leaden snow
i dont see how you would solve this without extra info
sonic dust
because in the answer key it didnt mention the answer to Q3c
anyway thanks for the help :D @leaden snow
cedar kilnBOT
Available help channel!
upper valley
how do I solve this?
3^(x-10)-2^(x-9)-3^(x-11)-2^(x-12)<0
cedar kilnBOT
how do I solve this?
3^(x-10)-2^(x-9)-3^(x-11)-2^(x-12)<0
wraith daggerBOT
Ansh
fallen heath
take $$t = \qty(\frac{2}{3})^{x - 13}$$ and you'll get a linear inequality in "t"
wraith daggerBOT
Ansh
pearl nest
Where is the point of A'(4, 0) in cartesian system?
cedar kilnBOT
Where is the point of A'(4, 0) in cartesian system?
pearl nest
Is that ' means something or its just a point
dire geode
Where is the point of A'(4, 0) in cartesian system?
show the original question
cedar kilnBOT
@pearl nest Has your question been resolved?
pearl nest
no
i will show the original question
I will translate for you
A(2a-1, 3b+2) point is in Q1 and its distance to x axis is 8, y axis is 5
so what is the geometric location of A'(a+1, b-2)
a)On the x axis A'(0,4)
b)On the y axis A'(4,0)
c)On Q2 A'(-3,5)
d)On Q3 A'(3,1)
e)On Q4 A'(1,-3)
magic spindle
How I can find b(h) or h(b) in Wolfran Alpha from
cedar kilnBOT
How I can find b(h) or h(b) in Wolfran Alpha from
graceful karma
Ask it to solve for one of the variables
solve (insert equation here) for (insert variable here)
,w solve xy+sqrt(x)=0 for y
wraith daggerBOT
muted widget
if $\cos (2x) = 1 - 2 \sin^2(x)$ is true, does that imply $cos (2Bx) = 1 - 2 sin^2(Bx)$ for dB/dx = 0 ?
wraith daggerBOT
madlor
cedar kilnBOT
@muted widget Has your question been resolved?
muted widget
not here, but yes
magic spindle
Any idea to solve ((xy)^(5/3))/((x+2y)^(2/3)) - 4.2/sqrt(0.013) * sqrt( ((0.013)^2x+2(0.035)^2*y ) / (x+2y)) = 0 for x or solve for y?
magic spindle
muted widget
sure as hell is not a pretty one
magic spindle
You cant just copy and paste in Wolfram, but I haven't been able to isolate any of the terms satisfactorily
muted widget
only thing i can really think of is maybe replacing x+2y=A lol
then the thing multiplied with the root, call it B
makes it less of a mess to look at
can insert them back later i reckon
glad kestrel
what exactly is the og question
magic spindle
I just want to isolate x=blabla or y=blublu of this equation
I've done this a few times in Wolfram, but I couldn't for this function
I already saw that there is a function x(y) or y(x) for y>0 or x>0.
My only idea was to consider some points (x,y(x)) by the above equation and try to build the function via interpolation
But I don't know, I have the intuition that there is a (very ugly) explicit function for x(y)
or y(x)
😢
cedar kilnBOT
@magic spindle Has your question been resolved?
eternal comet
why do you want an explicit y and x expression?
magic spindle
Well
I have this function g(x,y)=x+2y that want to minimize
For this i need to write g as g(x) or g(y)
magic spindle
why do you want an explicit y and x expression?
Like g(x)=x+2y(x) or g(y)=x(y)+2y
I thought about using lagrange multipliers, but I don't know if it works.
cedar kilnBOT
@magic spindle Has your question been resolved?
cedar kilnBOT
@magic spindle Has your question been resolved?
cedar kilnBOT
@magic spindle Has your question been resolved?
cedar kilnBOT
@magic spindle Has your question been resolved?
fallen heath
You probably do not want to solve that one lol
Give me the original question
iirc, that was a given function and you had to minimize the other function, on it
marble latch
cedar kilnBOT
marble latch
This uses arithmetic and geometric progression btw
wet oriole
write p & q in terms of each other
chilly warren
So far it's good. Now what the arithmetic progession says about p and q?
marble latch
So far it's good. Now what the arithmetic progession says about p and q?
I’m not sure how to incorporate the fact that it’s third and fifth months
chilly warren
You replace the n
marble latch
With 3 and 5?
chilly warren
yes
marble latch
chilly warren
Also, you seem to use one-based index
marble latch
Also, you seem to use one-based index
Is that ok?
chilly warren
Ok
marble latch
Ok! Ty
chilly warren
Why can't q be -4?
marble latch
Cause it’s books
chilly warren
Oh, I thought it's money, which means minus represent losses
hasty pelican
Sure
marble latch
So the answer be +-(3/2)?
hasty pelican
So the answer be +-(3/2)?
Yes, but it would be weird
Since it’s the sales of books
And having a negative common ratio results in every odd term being negative (-1^odd number = -1)
cedar kilnBOT
hasty pelican
No problem 
cedar kilnBOT
Available help channel!
marble latch
cedar kilnBOT
hasty pelican
Just a minute
Okay
So 31 is the highest stack
And it reduces by 2 every “term”
So another ap
hasty pelican
No no
We know n
We need to find the sum of the chairs
According to the question
marble latch
Oh
hasty pelican
Do you understand?
marble latch
marble latch
Oh
hasty pelican
Read the question again
marble latch
I thought we multiply by 12 afterwards
hasty pelican
12 stacks of chairs
hasty pelican
I thought we multiply by 12 afterwards
That would work if each stack had the same number of chairs
But it is decreasing according to the question
marble latch
Uhuh
hasty pelican
I’ll be back in a few minutes
We’re dealing with number of chairs not the height of the chair by the way
hasty pelican
Hey
hasty pelican
Sry i went to take my blood pressure just now HAHAHA
Oh, hope you’re doing good
hasty pelican
Yes, so the thing is, we need to find the total amount of chairs
And there is no need for 300 because we already know the maximum number of chairs
Which is 31
hasty pelican
a would be 31
Because that’s what it starts from
The series is something like: 31, 29, 27, 25, 23… because the stack reduces by 2
marble latch
D would be -2 right
hasty pelican
Yes