idk about "easier", but you can do it any way you want
#help-39
viscid shale
rare holly
how abt this we are given %
carmine schooner
u can verify a solution is correct by using concrete values for sure, but it's best to keep magic numbers out of ur solving in the long run
autumn fossil
how abt this we are given %
percentages are equvialent to ratios more or less
so the same thing holds
autumn fossil
If you were given ratio of volume to side or sth like that, then you wouldnt be able to pick
viscid shale
most geometry problems where you are only given relations and / or angles
autumn fossil
they have to be "homogenous" in the sense that you are comparing the same thing (e.g. length : length, area : area or volume : volume)
viscid shale
This is a really common example.
ABCD is a square with sidelength a with a quarter circle inside it.
Whats the ratio of the shaded area to the square's area.
In these cases we would choose a = 1 to simplify the whole problem
viscid shale
it mostly relies in algebraic intuition
And because enlisting all possible cases would make an absurdly long list.
rare holly
This is a really common example.
`ABCD` is a square with sidelength `a` with a q...
why in here we are allowed to assume a=1
viscid shale
Being asked about a ratio between two areas
In this case, the two scale in the same manner
If i choose a side length of 2, then both areas would be four times as before, and so, the ratio stays equal.
rare holly
I see
rare holly
>Being asked about a ratio between two areas
but if you choose 1 it would be 1x1=1 but if u choose 2 is 2x2 whihc 4
how does it stay same
viscid shale
Because the area of the circle will be four times the previous number too.
Also, its pretty easy to prove that the ratios stay equal.
rare holly
how
viscid shale
if you dont mind, id like to show you a simple method to solve your own case
the volume problem
rare holly
yh ok
rare holly
the question i got wrong was in the test there was a length x and we had to assume sm value i wish i rememberd it
rare holly
do you agree that 50% more means 150% * the previous amount
yh
viscid shale
and that 150% is equal to 3/2?
rare holly
3/2=1.5
viscid shale
yep, multiply by 100 to get the % and its the same.
rare holly
yh
viscid shale
well, then we have the following two equations
jolly parrotBOT
viscid shale
where P, Q and R refer to the volumes
rare holly
yh
viscid shale
Reading from the first line, you can "input" the value of Q into the 2nd equation
$R = \frac32 \left(\frac32 P\right)$
jolly parrotBOT
viscid shale
$R = \frac94 P$
jolly parrotBOT
rare holly
yh
viscid shale
this is telling you the volume of R as a fraction of P
divide both sides by 9/4 to cancel
and you get
$\frac 49 R = P$
jolly parrotBOT
rare holly
yh
viscid shale
and thats it, again, i made it intentionally long
but this method is far faster than choosing 100 as reference too.
rare holly
i see there are other ways to solve
rare holly
but this method is far faster than choosing 100 as reference too.
thanks man
viscid shale
I suppose youre in lower education rn, but being able to do this kind of operations will come useful later down the line
pearl pondBOT
pearl pondBOT
Available help channel!
unborn charm
when this was discussed in class my instructor said for part ii we'd need a couple more rules such as the following, 1 is the identity aka 1 g = g for g in the set, similarly -1 g = -g = g (-1) so on and assumed associativity and only then proceeded, then she went on to make the remark that once associativity is assumed we could have used a different set of rules and even then the multiplication table would look the exact same, why did she say that?
unborn charm
no like she said we could have used other asspciativity relations
still get the same table
plush bramble
you basically want us to read her mind
"we could have used a different set of rules" is pretty vague
unborn charm
different associativity relations
tropic saddle
what is that even supposed to mean. either the operation is associative or it isnt
unborn charm
she assumed associativity only on a set few elements, for example whilist calculating ik or jk for example
ik = iij = -j , and then she said it can be argued associativity holds throughout
that is, for any 3 arbitrary elements
tropic saddle
well ok if you assume enough associativity conditions then sure at some point you get associativity for the whole group
and there are different minimal assumptions you can make
unborn charm
and there are different minimal assumptions you can make
how would I know that
like why is this true
tropic saddle
well seems rather unlikely that the opposite is true
that somehow there is only one set of assumptions from which all other ones follow
but thats of course not a proof
I dont actually know whether its true
pearl pondBOT
pearl pondBOT
Available help channel!
daring bay
pearl pondBOT
daring bay
did i do it correct or
i wouldve asked chat gpt but i dont think it can recognize my writing
the a)
fresh olive
,rccw
jolly parrotBOT
fresh olive
For the first part you should be using binomial distribution
daring bay
~~For the first part you should be using binomial distribution~~
but it says to use the normal formula
above
in the text
fierce sky
Then use the normal approximation to the binomial distribution
daring bay
i did but
im not sure if its correct because when i use my calculator to see if its correct
the binomial result gives another value
fierce sky
What value did you get for the normal approx?
daring bay
What value did you get for the normal approx?
0,1064
is also weird that sigma is exactly 3
daring bay
fresh olive
,rccw
jolly parrotBOT
fresh olive
Your 1's look a lot like 7's
fresh olive
You can't put the k into the density formula and use the output
Density isn't probability
daring bay
You can't put the k into the density formula and use the output
what bro
wait
i literally copied the formula
fierce sky
Depends on your curriculum, but you probably need to use continuity corrections
Which is taking a CDF from 7.5 to 8.5
fresh olive
I suppose you could use density as a probability, but its an approximation :P
jolly parrotBOT
fierce sky
I suppose you could use density as a probability, but its an approximation :P
It's close enough
with continuity corrections I get 0.1062
fresh olive
I mean we are already approximating
fierce sky
with Pdf I get 0.1064
daring bay
why is sigma exactly 3 this is contradictory to the rule
fierce sky
Binomial gives 0.114
fresh olive
I haven't done it this way before so I'll leave it to ppl who know this way
fierce sky
Wait
daring bay
Wdym by pdf
fierce sky
Probability distribution function
daring bay
oh
i have to do -0,5 +0,5 right
or is that in another case
when integrating or something
fierce sky
Honestly depends on your curriculum
They give pretty close values anyways
But yeah I would use continuity corrections
daring bay
But yeah I would use continuity corrections
i gotta do that
fresh olive
Probability distribution function
-# I thought it was probability density function
daring bay
because the value isnt height or something right
thats when u gotta do the corrections?
fierce sky
why is sigma exactly 3 this is contradictory to the rule
np ≥ 5, n(1 - p) ≥ 5 in this case are satisfied
fierce sky
thats when u gotta do the corrections?
Well because taking a point value isn't always accurate since the normal distributions are continuous
fresh olive
-# Wait actually there's both probability density function and probability distribution function
fierce sky
I guess that's the motivation for taking continuity corrections? Although there may be other reasons I'm not aware of
fierce sky
-# Wait actually there's both probability density function and probability distr...
It should be the density function
daring bay
I guess that's the motivation for taking continuity corrections? Although there ...
i thought u use it when you got stuff like coin tosses, 2 sorts of balls
and you dont use it when u got heights, and all that
fresh olive
-# My best guess is distribution includes density and mass functions
fierce sky
i thought u use it when you got stuff like coin tosses, 2 sorts of balls
Idk, you use it when you wanna approximate it with the normal
I don't see why else you wouldn't use binomial since that's the most accurate
daring bay
Bro im so confused wdym by approximate
fierce sky
You'll probably have to research more on why since I don't know
fierce sky
Bro im so confused wdym by approximate
You could have easily done this question with the binomial distribution
But we're using a normal approximation here
fresh olive
Approximate: Close to the actual value but likely not completely exact
daring bay
Approximate: Close to the actual value but likely not completely exact
oh ok
fresh olive
So approximation of pi is 3.14 or 22/7
daring bay
Yeah
Man lowkey this topic is confusing as hell
also for the b)
my n is 50 right
or is it 10
fresh olive
The trial here is pulling a ball out, which is done 50 times
daring bay
oops
daring bay
The trial here is pulling a ball out, which is done 50 times
yes
fresh olive
the first bit tells you 4/10 are red so probability p=0,4
fresh olive
$variance=500,40,6=12$\
$\sigma=\sqrt{variance}$
daring bay
what does the variance even mean in probability context
jolly parrotBOT
BBMaths
fresh olive
$$\text{Variance of }X=\text{Var}(X)=\sigma^{2}$$
jolly parrotBOT
BBMaths
fresh olive
Variance is very important in probability and statistics
fierce sky
It's a way to measure the distance of a datapoint from the average value
Or, how spread the distribution is
fresh olive
For two independent random variables, $X$ and $Y$,
$$\text{Var}(X+Y)=\text{Var}(X)+\text{Var(Y)}$$
jolly parrotBOT
BBMaths
fresh olive
standard deviation doesn't have this property
Standard deviation is typically defined as the sqrt of variance as well
daring bay
Standard deviation is typically defined as the sqrt of variance as well
yeah
the standard deviation is just how far the values goes off the expected value? or
actually im gonna watch a yt video about that in my own native language
fresh olive
$$\text{Var}(X)=\mathbb{E}[(X-\mathbb{E}[X])^{2}]$$
$$\sigma(X)=\sqrt{\mathbb{E}[(X-\mathbb{E}[X])^{2}]}$$
Not $\mathbb{E}[|X-\mathbb{E}[X]|]$ (Average distance from expected value)
jolly parrotBOT
BBMaths
daring bay
mu=np=50*0.4=20
btw in the b) the expecred value was 20, so its expected that you pull 20 red balls on 50 draws in context meaning?
fresh olive
Yeah
But the probability of pulling exactly the expected value goes towards 0% as n goes to infinity
daring bay
yeah
fresh olive
So if you had 1.000.000 balls and you drew 200.000 that would be unusual
daring bay
o
yep
ty
im gonna take a look at cumulative normal distribution then head over to hypothesis tests, the worst
fresh olive
You'd more likely get something like 201.347
The deviation grows with sqrt(n) so it would deviate 1.000 times more than just 1 ball
daring bay
is it possible to read out the standard deviation out of a histogram? or
fresh olive
Kind of
It will be an approximation since we don't know what the original values were
pearl pondBOT
@daring bay Has your question been resolved?
daring bay
It will be an approximation since we don't know what the original values were
yeah ty bro for everyhing
quick venture
pearl pondBOT
quick venture
can someone help me find d
fresh olive
You need to simplify
$$84*10^6=\frac{143.24\left(\frac{d}{2}\right)}{\frac{\pi}{2}\left(\frac{d}{2}\right)^4}$$
jolly parrotBOT
BBMaths
fresh olive
Multiply both sides by the denominator
pearl pondBOT
@quick venture Has your question been resolved?
quick venture
fresh olive
Now divide both sides by d/2 so you only have d on one side
quick venture
fresh olive
And then move the $8410^6\pi/2$ to the other side by dividing
jolly parrotBOT
BBMaths
quick venture
fresh olive
Then cube root both sides
quick venture
fresh olive
And then double both sides and you'll have $d=2*\sqrt[3]{\text{Something}}$
jolly parrotBOT
BBMaths
quick venture
fresh olive
The 2 times should be outside the cube root
If you bring it inside it becomes 8 times
fresh olive
You can keep simplifying
pearl pondBOT
fresh olive
But now you have d=something so that should be enough to use a calculator
pearl pondBOT
Available help channel!
rare field
pearl pondBOT
cursive wraith
heres where im jammed <:catthink:526983414226747393>
I don't think you need to worry about dimensions
rare field
wait really
cursive wraith
Take any element in $X\cap X^\perp$
jolly parrotBOT
Rafilouyear2026
rare field
hmmmm oke ill head in that direction and see
cursive wraith
I can drop a hint if you're still stuck
But the hint kinda solves everything so try before asking
jolly parrotBOT
Rafilouyear2026
cursive wraith
So $|v| = 0$, meaning $v = 0$
jolly parrotBOT
Rafilouyear2026
pearl pondBOT
pearl pondBOT
Available help channel!
ruby wharf
Hi I was hoping I could get some guidance here.
pearl pondBOT
Hi I was hoping I could get some guidance here.
ruby wharf
!status
pearl pondBOT
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
2
Or maybe 1
So Sal Khan before I paused the video (I got this puzzle if you will, from Khan Academy) mentioned that he saw 3 triangles, the two small ones and the big one that encompasses both.
Angle ABC is a right angle, and so is angle DBC
Also we can see that the big triangle has a side length of the sum of segments AD and DC, so AC should be 8.
fierce sky
You can use the fact that ABC and BDC are similar triangles to construct a relation between AC, DC and BC
ruby wharf
I think I should start by proving similarity where it exists, I know that triangle BDC and triangle BDA share side BD and that segment BD is congruent to itself by reflexivity.
ruby wharf
You can use the fact that ABC and BDC are similar triangles to construct a relat...
Thank you! Let me think why they're similar, I don't see how they are similar yet.
They do both have a right angle respectively. If we somehow knew one other angle and we knew that it was congruent to both triangles then we could prove their similarity.
Maybe I can split triangle BDC into two?
rare field
hmmm could we write out pythagoras for every triangle there then solve for BC?
fierce sky
Thank you! Let me think why they're similar, I don't see how they are similar ye...
They share two common angles
One of which is the right 90° angle
ruby wharf
hmmm could we write out pythagoras for every triangle there then solve for BC?
But we only know segment BC's length and segment AD's length, I think maybe we could if we knew the proportions/ratios and which sides corresponded to which sides and if the triangles are similar to each other?
ruby wharf
They share two common angles
Omg you're right!!!!
fierce sky
:)
ruby wharf
That's definitely an essential detail, but because BCD is the same angle for the big triangle and for triangle BDC, we also know that angle BAD is shared with triangle BDA and the big triangle ABC.
Right?
rare field
I think I've found BC with my method
ruby wharf
I think I've found BC with my method
Please don't tell me, I don't want the solution but just to think in a fun way
Also, I think there's multiple methods to get the same results maybe?
hidden hawk
I think I've found BC with my method
Law of cosines was my first guess too
rare field
Also, I think there's multiple methods to get the same results maybe?
yea with these geometry problems always 
dont worry im not saying
ruby wharf
Triangle BDC is similar to triangle ABC because of Angle-Angle triangle similarity.
hidden hawk
I was thinking maybe there was a nice construction based proof
ruby wharf
Triangle ABD is also similar to triangle ABC
But because we know that the two small triangles are similar to the big triangle, that means that the two smaller triangles are similar to each other as well because of something I forgot what it's called
How would you write that in a proof?
∆ABD~∆CBA~∆BDC
ruby wharf
How would you write that in a proof?
I mean the reason
Ah this is what I was looking for, I was thinking of the transitive property.
Is this used often in two column proofs?
pearl pondBOT
@ruby wharf Has your question been resolved?
pearl pondBOT
@ruby wharf Has your question been resolved?
turbid summit
I dont think Ive seen transitive property in 2 column proofs
ruby wharf
I dont think Ive seen transitive property in 2 column proofs
I have quite a few times in Khan Academy.
ruby wharf
Ah this is what I was looking for, I was thinking of the transitive property.
Which is also where I got this picture from @turbid summit
turbid summit
It could be regional differences, but the 2 column proofs here are revolved on similarity and congruence tests
So like AA, SAS, etc
ruby wharf
So like AA, SAS, etc
Yea, for example if triangle A is similar to triangle B, and triangle B is similar to triangle C, then triangle A is similar to triangle C by the transitive property.
turbid summit
Try finding BD first
ruby wharf
Try finding BD first
Ok, we know that the hypotenuse of triangle ABC is 8
fierce sky
just use the ratios between AC, DC, and BC
ruby wharf
Because every triangle's side that is opposite to the right angle is the longest side, and for triangle ABC that side has length 8
turbid summit
Correct
turbid summit
Whats the purpose of using the transitive property in the first place?
ruby wharf
How do i figure out the ratios though? I know that the base of triangle ABC relates to the number 6 and number 2, but I don't know by what proportion exactly
ruby wharf
Whats the purpose of using the transitive property in the first place?
I suppose if I know side AB for example, then I can know side BC
maybe
eh idk
I was just stating it because its something true
turbid summit
We know that by the property ADB is similar to BDC, right?
fierce sky
How do i figure out the ratios though? I know that the base of triangle ABC rela...
AC/BC = BC/DC
ABC_hyp / ABC_short = BCD_hyp / BCD_short
ruby wharf
just use the ratios between AC, DC, and BC
Hmmm
ruby wharf
We know that by the property ADB is similar to BDC, right?
exactly
turbid summit
So based on the sides of the triangles, we'll use their ratios
Because they are scaled the same
So for example hyp/hyp=adj/adj
ruby wharf
AC/BC = BC/DC
Is this correct?
fierce sky
:)
ruby wharf
But we also know what DC is
True!
i forgot that XD
ty for pointing that out 1 second
YAYYYY THANK YOU ALL SOOOOO MUCH!!! 🩷
This was so cool to solve!!!
turbid summit
Of course there are many other ways, but this is the shortest
ruby wharf
Of course there are many other ways, but this is the shortest
Interesting
turbid summit
I was thinking of using pythagoras theorem
ruby wharf
I was thinking of using pythagoras theorem
Same!
idk how that'd be possible though without like maybe laws of sines etc?
i haven't learned those yet
again, much appreciated! 🩷
turbid summit
idk how that'd be possible though without like maybe laws of sines etc?
Its possible
Because of ADB and BDC
ruby wharf
Its possible
interesting
ruby wharf
Because of ADB and BDC
Could you set the length of BD to be anything you want?
turbid summit
No
ruby wharf
oh
ruby wharf
Oooooo
Ty for helping me I sent you both friend requests if that's ok and not weird 🩷
turbid summit
Thats fine
pearl pondBOT
still hamlet
<@&268886789983436800>
limber dirge
Ah this is what I was looking for, I was thinking of the transitive property.
YO STOPPP
ruby wharf
Are you in geomatry are calculus?
Neither, I finished Algebra 1 in 9th grade so far but I'm self-learning stuff all across math right now.
Let's move to #discussion
limber dirge
ok
pearl pondBOT
Available help channel!
limber dirge
Hi
left steppe
Can someone CLOSE?
do .close
limber dirge
.close
pearl pondBOT
limber dirge
.reopen
pearl pondBOT
✅ Original question: #help-39 message
proper nova
!status
pearl pondBOT
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
It is VERY confusing
proper nova
also which question btw
limber dirge
also which question btw
1-4
limber dirge
what have you tried for `1`
I looked at the graph but cant find the exact value
does t stand for time?
proper nova
I looked at the graph but cant find the exact value
so when people board the roller coaster
proper nova
I looked at the graph but cant find the exact value
so try to think
when people are boarding the roller coaster
that means the roller coaster is not running
limber dirge
so 0 seconds?
proper nova
yes
limber dirge
OH SO I JUST PLUG IN 0
proper nova
life skills issue 
proper nova
yeah lol
limber dirge
life skills issue <:opencry:586078614865444865>
for number 2 I estimate 2 seconds
limber dirge
so can I use any 0
proper nova
but before we move on to 2, can you see what's the goal for this
they mentioned in the exercise
limber dirge
they mentioned in the exercise
factor?
proper nova
factor?
yes!
and when the roller coaster is below the ground
it is when $h(t) < 0$
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
so try to factor $h(t)$ and see what you get
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
with this kind of polynomial, it's best to find the roots of $h(t)$ first
jolly parrotBOT
1 divided by 0 equals Infinity
limber dirge
with this kind of polynomial, it's best to find the roots of $h(t)$ first
so I know 1 root is 2 now I need to find other
proper nova
because $\forall f(x)$ where $f$ is a polynomial, $f(x) = (x - r)g(x)$ where $r$ is the root of $f(x)$
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
so I know 1 root is 2 now I need to find other
that's kinda complicated
so if you know a root is 2
just factor anyway
limber dirge
ok
I used synteticc divison
I divided by 2
and got remainder 0
so its a zero?
proper nova
there are odd numbers is here, are you sure you want to divide by 2?
proper nova
mhm
limber dirge
4t^2-13t-17
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
so let's check the conditions for $h(t) < 0$
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
more precisely, what does factoring has to do with this
proper nova
**1 divided by 0 equals Infinity**
factoring helps us split $h(t)$ in to factors
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
specifically 2 factors
proper nova
cool then
proper nova
**1 divided by 0 equals Infinity**
let's keep this for simplicity now
3 factors are going to get complicated af
limber dirge
ok
proper nova
**1 divided by 0 equals Infinity**
now when does this thing < 0
(hint, consider the signs of the factors)
limber dirge
wont be = to 0
proper nova
that's kinda obvious lmao
try considering the signs of the factors
simpler
when is $A \cdot B < 0$
jolly parrotBOT
1 divided by 0 equals Infinity
limber dirge
try considering the signs of the factors
when its x 0
Bro I am so confused what u telling me
why are there inequalities
why cant u just get 3 factors and set all of them to 0
and get -1, 2 , 17/4
why inequalities?
proper nova
**1 divided by 0 equals Infinity**
this happens when $A < 0$ and $B > 0$ or $A > 0$ and $B < 0$
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
talking about inequalities, this is called making a sign table
limber dirge
talking about inequalities, this is called making a sign table
This is algebra 2
math
proper nova
yo I dont think I am suppose to do sign tables
for 3 factors, you gotta make a table to keep track of the signs
limber dirge
ur prob doing calculus 3
proper nova
for 2 factors
proper nova
**1 divided by 0 equals Infinity**
this is enough
limber dirge
this is enough
Yo I never learned this in my life track of the signs
proper nova
lmao
limber dirge
is this what u learn in china?
proper nova
i think it's basic algebra
proper nova
is this what u learn in china?
yo im not chinese
limber dirge
*yo im not chinese*
ur from Japan?
limber dirge
ok
proper nova
**1 divided by 0 equals Infinity**
this is algebra basic
proper nova
**1 divided by 0 equals Infinity**
so anyways, these conditions can be applied here
:fire
limber dirge
its called sign table?
proper nova
its called sign table?
sign table is a table of signs
but for now, we are just stating algebra basics
multiply 2 numbers with opposite signs gives negative values
limber dirge
multiply 2 numbers with opposite signs gives negative values
So a zero is confirmed where the sign changes.
proper nova
So a zero is confirmed where the sign changes.
🔥
proper nova
**1 divided by 0 equals Infinity**
anyways, we can apply conditions we got here
so split into cases
- $\begin{cases}
t - 2 < 0\
4t^2 - 13t - 17 > 0
\end{cases}$
jolly parrotBOT
1 divided by 0 equals Infinity
limber dirge
I am so co confused on sign tables bec I never learned it
why is t-2 less then 0
?
proper nova
I am so co confused on sign tables bec I never learned it
you don't need to learn that for you
proper nova
so split into cases
@limber dirge
limber dirge
<@956611822365057064>
yes
proper nova
we split into cases that can happen
proper nova
**1 divided by 0 equals Infinity**
so one case is when this happens
and other case is when $\begin{cases}
t - 2 > 0\
4t^2 - 13t - 17 < 0
\end{cases}$
jolly parrotBOT
1 divided by 0 equals Infinity
limber dirge
so these 2 cases can happen
limber dirge
we solve for those 2 cases
ok
proper nova
Is this descrates rule of signs?
i don't really know
proper nova
**1 divided by 0 equals Infinity**
anyways
for this case
$t - 2 < 0$ then we can solve $t < 2$
$4t^2 - 13t - 17 > 0$ then we are going to use the factorization you did earlier
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
$(4t - 17)(t + 1) > 0$, apply the same conditions
jolly parrotBOT
1 divided by 0 equals Infinity
pearl pondBOT
@limber dirge Has your question been resolved?
pearl pondBOT
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formal radish
number of solutions of the tan x + sec x = 2 cos x lying in the interval [0,2pi] is
formal radish
are the solutions 5pi/6 and pi/6 or 3pi/2 and pi/6
?
ik one sol is pi/6 whats the other
uneven cypress
What have you done?
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
toxic lichen
how did you get either one of these
formal radish
i got sin x =1/2 and sinx = -1
uneven cypress
i got sin x =1/2 and sinx = -1
How did you come to that
But if that is correct then all three of the aforementioned angles are solutions
formal radish
but the answer shows 2 angles
uneven cypress
Maybe show us your work then?
formal radish
uneven cypress
You also need to account for cos x not equal to 0
Otherwise multiplication on both sides does not give an equivalent equation
And in general the tangent and secant values have to be defined
vernal bear
Right
pearl pondBOT
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bitter herald
Is it the case that for two band-limited functions $g_1(t)$ and $g_2(t)$ with corresponding bandwidth of $W_1$ and $W_2$ respectively, the bandwidth of $g_1(t)\2g_2(t)$ will be $W_1+W_2$?
jolly parrotBOT
bitter herald
Actually, I think it is the case that
\begin{align*}
g_1(t)\2 g_2(t) &\Implies \txs{BW:} W_1+W_2 \
g_1(t)+g_2(t) &\Implies \txs{BW:} \max(W_1,W_2) \
g_1(t)g_2(t)&\Implies \txs{BW:} \min(W_1, W_2)
\end{align}
jolly parrotBOT
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slim moss
pearl pondBOT
slim moss
i have a question about this
if im not mistaken in my calculation it should be lim n -> +inf of (e^x/n!)
but wouldnt that also be +inf/+inf
autumn fossil
if im not mistaken in my calculation it should be lim n -> +inf of (e^x/n!)
why n!?
slim moss
if you took the limit
autumn fossil
so n! just some finite number (though possibly big)
pearl pondBOT
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blissful salmon
This is a weird question… Can anyone try to compute a value of n other than 2 such that the binary representation of 3^n+1 contains exactly 2 1’s, and can provide an example as to how 3^n+1 can be expressed as the sum of two powers of 2?
autumn fossil
its (3^n) + 1, right?
blissful salmon
its (3^n) + 1, right?
Yes
autumn fossil
i tried all powers up to a hundred and there are none, it seems like the n.o. ones is rising
(though its not strictly increasing)
3^n + 1 = 2^a + 2^b
you'd need this
blissful salmon
you'd need this
Yes
blissful salmon
i tried all powers up to a hundred and there are none, it seems like the n.o. on...
Yes, it seems to rise
autumn fossil
mod it by 4
oh nvm that doesnt wokr
modding it by 8 could though
3 -> 1 -> 3 -> 1....
this is 3^n mod 8
so the LHS is 2, 4, 2, 4,...
which means either a or b is 1 or 2
naive zinc
Assuming a<=b then a has to be 1
unborn abyss
which means either a or b is 1 or 2
well it's definitely not 2^a + 1
autumn fossil
well it's definitely not 2^a + 1
i meant 2^a + 2^1 or 2^a + 2^2
naive zinc
Since a,b are odd by mod 3 argument
viscid shale
well, to start with, 3^n + 1 is always even*** so it always has the least significant bit = 0 at all times
autumn fossil
so we have 2^b = 3^n - 1
subtract 1 and you get 2 mod 8
autumn fossil
oh rightt
naive zinc
3^m+1, 3^m-1, one being 2 another being 2^(b-1)
autumn fossil
so it seems like mod 8 + some algebra
blissful salmon
Alright, thanks for everyone’s help. Now here’s the next question. Does there exist positive integer pairs (m,n) other than when m=n or when n is 1 or 2 such that (3^n-2^n)/(2^m-3^n) is also an integer?
autumn fossil
have u tried anything already?
blissful salmon
have u tried anything already?
Not really… idk where to start
autumn fossil
id start by making it diophantine
fractions in NT arent usually helpful
make it a simple equation with no fractions, possibly at the cost of adding a variable
blissful salmon
id start by making it diophantine
Yes, so I already tried to let (3^n-2^n)/(2^m-3^n) = X for X is an integer. And rearranging gives us 3^n-2^n = X(2^m-3^n) which leads to (X+1) * 3^n = X*2^m + 2^n. I figured we could probably factor the largest power of 2 from the RHS and know that X+1 is divisible by this largest power of 2, but idk where to go from there
naive zinc
(2^(k)-1)/(2^(k)-(1.5)^n) is an integer. Where k=m-n. Should eliminate cases
blissful salmon
(2^(k)-1)/(2^(k)-(1.5)^n) is an integer. Where k=m-n. Should eliminate cases
How did you arrive at this?
Idk how to approach the diophantine though other than what I mentioned before
autumn fossil
yeah, maybe it wasnt a good idea
autumn fossil
(2^(k)-1)/(2^(k)-(1.5)^n) is an integer. Where k=m-n. Should eliminate cases
this looks like dividing both numerator and denominator by 2^n, except the numerator is divided incorrectly?
or maybe im missing sth
oh i am missing sth
i got it now
$\frac{3^{n}-2^{n}}{2^{m}-3^{n}}=\frac{3^{n}-2^{m}+2^{m}-2^{n}}{2^{m}-3^{n}}=-1+\frac{2^{m}-2^{n}}{2^{m}-3^{n}}$
jolly parrotBOT
MathIsAlwaysRight
autumn fossil
so $\frac{2^{m}-2^{n}}{2^{m}-3^{n}}$
blissful salmon
m=2 and n=1 ?
autumn fossil
**MathIsAlwaysRight**
if m > n, then numerator > denominator
blissful salmon
**MathIsAlwaysRight**
Alright so we know that this has to be an integer
There might be a possibility that there exists nontrivial solutions to the Diophantine equation that I made…
Is it possible of checking by code?
pearl pondBOT
@blissful salmon Has your question been resolved?
pearl pondBOT
@blissful salmon Has your question been resolved?
blissful salmon
Alright, thanks for everyone’s help. Now here’s the next question. Does there ex...
This is the new question. My original question has been solved
blissful salmon
I happened to find that a very similar question was asked in mathoverflow that claims that the answer is no. I don’t really get their answer though: https://mathoverflow.net/questions/69700/are-there-any-solutions-to-frac3n-2n2k-3n-n?rq=1
fresh olive
Does there exist $(m,n)\in\mathbb{N}^{2}$ with $m\ne n$ and $n\ge3$ such that: $$2^m-3^n|3^n-2^m$$
blissful salmon
Does there exist $(m,n)\in\mathbb{N}^{2}$ with $m\ne n$ and $n\ge3$ such that: $...
Yes that’s my question
fresh olive
Actually the fraction might be better way of saying it since I'd recommend making an integer k with LHS*k=RHS
$$2^m-3^n=k3^n-k2^m$$
$$(k+1)2^m=(k+1)3^n$$
So k-1 is a power of 2 and k+1 a power of 3?
blissful salmon
I happened to find that a very similar question was asked in mathoverflow that c...
It would be great if someone could explain the first answer provided here. And I would like a link to de Weger’s thesis as mentioned in the answer
I think you factorised wrong
jolly parrotBOT
BBMaths
blissful salmon
No you still factorised it wrong
fresh olive
Oh I put k on wrong side :P
jolly parrotBOT
BBMaths
fresh olive
Still not right :/
Damn you might be having a better time than me at this problem
Oh wait is the numerator 3^n-2^n
blissful salmon
Oh wait is the numerator 3^n-2^n
Yes
It’s okay, at least I found the answer to the problem but I still need to know how they got the answer
blissful salmon
I happened to find that a very similar question was asked in mathoverflow that c...
The answer is in the link I provided
fresh olive
The 9 upvoted one?
$$2^m-3^n|3^n-2^n$$
$$2^m-3^n|2^{m-n}-1$$
$$2^m<3^{n+1}$$
$$m<(n+1)\frac{\log(3)}{\log(2)}$$
$$|2^m-3^n|<2^{(n+1)\frac{\log(3)}{\log(2)}-n}$$
$$|2^m-3^n|\ge\text{min}({2^m,3^n})^{0.9}$$
jolly parrotBOT
BBMaths
blissful salmon
It’s okay, thanks for trying though!
I think I’ll close this channel for now
.close
pearl pondBOT
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dense iris
pearl pondBOT
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@dense iris Has your question been resolved?
celest heath
your third step where you wrote probab of d/non spam = 0.05 is incorrect imo coz, it should be probab non detected/spam
nocturne grail
your third step where you wrote probab of d/non spam = 0.05 is incorrect imo coz...
you claimed a channel.
celest heath
huh
nocturne grail
just- close this and if jamie comes back
let them claim it and you can talk about it there
celest heath
just- close this and if jamie comes back
how to
nocturne grail
typing .close
celest heath
.close
pearl pondBOT
nocturne grail
🤩
celest heath
noice
dense iris
.reopen
dense iris
your third step where you wrote probab of d/non spam = 0.05 is incorrect imo coz...
wait can u explain that more
celest heath
sure
dense iris
I thought I was just notating down the info they gave saying that a false positive is 5 percent
frank violet
sure
It hasn't reopened yet, if you would love to explain, reopen it before it's locked
celest heath
.reopen
pearl pondBOT
✅ Original question: #help-39 message
celest heath
.close
pearl pondBOT
celest heath
.reopen
pearl pondBOT
✅ Original question: #help-39 message
celest heath
try expanding (A^-1 + B^-1)^5 using binomial theorem
how does that click in your brain
bronze heath
how does that click in your brain
-# avg jee shit
celest heath
yes
wraith hatch
what you probably take for granted is the commutativity
thank the lords for they have blessed you with diagonal matrices
celest heath
||What a lazy "+2026" to make 20XX answers appear||
pretty fair considering this question was formulated that way not by some organization but by an inexperienced teacher
celest heath
thank the lords for they have blessed you with diagonal matrices
that's thepoint, when I see matrices I stop thinking about normal variables to use in algebra
if the same was written in a and b i would have saw it
wraith hatch
what i meant was the reason u can combine terms into powers like A^3B^2 is because diagonal matrices commute under multiplication
young shore
random ermine
what i meant was the reason u can combine terms into powers like A^3B^2 is becau...
this is PRECISELY becuase multiplication of real numbers is commutative 🤓🤓
celest heath
.close
pearl pondBOT
celest heath
what i meant was the reason u can combine terms into powers like A^3B^2 is becau...
i understood that, thanks for the help 😁
pearl pondBOT
Available help channel!
viscid shale
,tex \def\e{\varepsilon}
Under the algebra of Dual Numbers, i.e. $w = a+b\e: \e^2 = 0, \e \neq 0$
Let's say I have a function $f$ that's differentiable. From the usual definition of the derivative we have: $$f'(x) = {f(x+\e) - f(x) \over \e}$$
Which can be easily manipulated to construct: $$f(x+\e) = f(x) + f'(x)\e$$
But now, and this is my question, we can't actually do a substitution of the form $x = y +\e$, right?
jolly parrotBOT
viscid shale
The 2nd equation ties nicely to the definition of f(x+ne) using taylor series.
But i think this particular way of constructing the identity is inherently flawed given the nature of the derivative here.
rustic atlas
you can't really divide by epsilon no?
viscid shale
yeah, that was mostly my concern.
toxic lichen
i think this works at most for functions already expressible as power series
viscid shale
but given the nature of epsilon it still makes somewhat sense, idk
i ended up with this identity, wait a sec
,tex \def\e{\varepsilon}
\begin{align*}
\e\sum_{k=a}^b f(x+k\e) &= \e\sum_{k=a}^b \big[ f(x) + \e\sum_{r=0}^{k-1}f'(x+r\e) \big] \
&= \e(b-a+1)f(x) + \e^2\sum\sum f'(x+r\e)\
&= \e(b-a+1)f(x)
\end{align*}
lazy tex ik
jolly parrotBOT
viscid shale
Still seems wrong
toxic lichen
nah it's correct
you're multiplying by epsilon so that the former real-number parts (which are all just f(x) each) get dumped into the epsilon part, and whatever former epsilon part gets dumped into nothingness
viscid shale
uh, ig. Im somewhat surprised that this whole construction which base was totally unfounded turned out to be true
thanks
.close
elfin stirrup
Does one really need the assumption that phi is a characteristic function?
elfin stirrup
.close
pearl pondBOT
pastel umbra
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Available help channel!
proper nova
I needed help with advanced geometry. Question: Given $\triangle ABC$ is circumscribed inside circle $(O)$ and $\triangle ABC$ is acute. $(O)$ has a diameter $AS$. Heights $AD$, $BE$, $CF$ of $\triangle ABC$ intersects at $H$. Let $I$ be the midpoint of $AH$, $M$ be the midpoint of $BC$. $EF$ and $AD$ intersects at $K$, $L$ is the intersection of $AS$ and $BC$
-
The circle that circumscribes $\triangle ABC$ intersects with $(O)$ at $N \neq A$. Prove that $M, H, N$ is collinear
-
Prove that $KL$ // $HM$
-
The bisector of $\angle BAC$ intersects $IM$ and $HM$ at $G$ and $T$, $GH$ intersects $AB$ at $X$. Prove that $\angle AXT = 90$ degrees
jolly parrotBOT
1 divided by 0 equals Infinity
proper nova
for anyone who wants to !status me
1
i can't even think how i would do 1
just a side note: euclidean geo and not algebraical geo
pearl pondBOT
@proper nova Has your question been resolved?
proper nova
<@&286206848099549185>
pearl pondBOT
@proper nova Has your question been resolved?
proper nova
Lmao
broken timber
Maybe try power of a point
proper nova
Maybe try power of a point
Pardon?
broken timber
This looks like power of a point spam
cyan briar
i guess my figure is wrong please chek it out
proper nova
This looks like power of a point spam
I haven't heard of a power of a point before
broken timber
You can search it up, I can try to explain iyw
proper nova
You can search it up, I can try to explain iyw
Aight
proper nova
You can search it up, I can try to explain iyw
Hm
broken timber
So given a circle and a point, draw two lines that intersect the circle twice
The formula is then (point to first intersection * point to second intersection) is always the same
proper nova
broken timber
Yes
proper nova
You're telling me that we spam those
broken timber
You can find the other lengths
proper nova
I needed help with advanced geometry. Question: Given $\triangle ABC$ is circums...
I think that power of a point thing should be used on the circumcircle that circumscribes AEF
broken timber
S could be the point then
proper nova
SH.SN=S(something).SA
broken timber
where something is SO I think
proper nova
If im trying to prove that M, H, N is collinear then i need to prove ASM = ASH = ASN
proper nova
where something is SO I think
I don't think so
broken timber
Oh ok
proper nova
If im trying to prove that M, H, N is collinear then i need to prove ASM = ASH =...
How am i going to relate this, to the power of points
broken timber
I’m not sure yet
proper nova
Well i can see that ANS = 90
broken timber
Yes
cyan briar
YK WHAT I AINT EVEN TRYING IM SORRY
frank violet
I needed help with advanced geometry. Question: Given $\triangle ABC$ is circums...
Show N,H,S are collinear
Show BHCS is parallelogram
=>done
pearl pondBOT
@proper nova Has your question been resolved?
acoustic narwhal
1/0 is a division against the horizon
Not only would it not resolve into a finality
But it would be something that goes unbound to create a new status
“If sally divided one apple to no one”. She would still have 1 apple.
And for her to define NOone she would have to compare it to someone
Meaning 0 cannot be computed unless a computation that explained 0 exists
0 is a status
Not a value
there, now no one can make a better argument!
calm wing
1/0 is a division against the horizon
Not only would it not resolve into a fina...
brother 🥀
ashen ivy
1/0 is a division against the horizon
Not only would it not resolve into a fina...
brother 🥀
acoustic narwhal
Youre welcome (;
fresh olive
1/0 is a division against the horizon
Not only would it not resolve into a fina...
brother 🥀
That's their name lol
acoustic narwhal
I just showed up, saw it and thought it was funny to throw my hat in, I have no context here lol
fresh olive
The name of the person who needs help is called 1/0=infinity
fresh olive
Look at the pinned message for the problem
daring bay
hello so for the a) , using the fi and the table method
pearl pondBOT
hello so for the a) , using the fi and the table method
daring bay
the result is 0,033 but you have to do 1 - 0,033
why? it hasnt been like this till now in normal distribution
pearl pondBOT
@daring bay Has your question been resolved?
mystic geode
the result is 0,033 but you have to do 1 - 0,033
Sorry can you explain how you got this
daring bay
Sorry can you explain how you got this
Chat gpt said that
do u want to see what i calculated
rq
?
mystic geode
ye
mystic geode
,rotate
jolly parrotBOT
mystic geode
,calc 0.514*50000
jolly parrotBOT
Result:
25700
,calc sqrt(0.514*50000 ( 1-0.514))
jolly parrotBOT
Result:
111.75956334918
And we want Pr( 25500 ≤ X < 2600)
mystic geode
,calc (25500 - 25700)/111.76
jolly parrotBOT
Result:
-1.7895490336435
Ok but generally you dont have to do that 1 - result bullshit at the end right
mystic geode
So it is $$\phi(2.68) - \phi(-1.79)$$
daring bay
that scared me
jolly parrotBOT
InterGalactic
daring bay
**InterGalactic**
Oh
mystic geode
👍 ye
mystic geode
wdym
daring bay
wdym
Like
In this normal distribution thing
i never have to take the complementary event at the end?
so do 1 - Pr( 25500 ≤ X < 2600)
yeah
i was confused
mystic geode
When you have a greater something
like Pr(Z > a)
then you take the complemetary
So, Pr(Z>a) = 1- \phi(a)
oh wait there is phi of -1.79
Table has only positives
So you take complemetary
daring bay
So you take complemetary
hhhhhh
so i can just memorize, whenever theres a negative,
then i take the complementary?
mystic geode
Pr(Z < -a) = 1 - \phi(a)
mystic geode
then i take the complementary?
ye
mystic geode
ye
daring bay
bet tyyy
mystic geode
,w Z value of 1.79
jolly parrotBOT
daring bay
,w Z value of 1.79
Oh btw Z is just the value that the phi shit calculates
last bear
ok well its 3am and im in bed rn after working on this for school all day because im bad at math and was sick for 2 weeks but does anyone generlaly know what I should do here? I tried the thing with V(r, h) = the cylinder volume and then set up another equation to get r or h and then set that thing zero but I have no idea what to do to get like the ratio of the volumes so maybe if someone could give me a quick overview of how to go about solving itd thatd be nice so i can give it another shot tommorow
A sphere of radius R is a rotating cylinder of greatest volume
to be registered (an axis section is shown opposite)
Determine the ratio of the volumes of the two bodies to each other!
pearl pondBOT
ok well its 3am and im in bed rn after working on this for school all day becaus...
@last bear Has your question been resolved?
Available help channel!
exotic gale
The area of the region enclosed by |x+2| + |y-a| =b is 20 units^2. The product ab is 30. What is the value of a? Express youur answer in simplest radical form.
exotic gale
i think set x=-2 and rewrite |y-a| as a piecewise function
acoustic plaza
For x>-2 the first modulus will open with a positive sign and then plot |y-a| = b - x - 2
Then for x<-2 plot |y-a| b + x + 2
exotic gale
wdym modulus
acoustic plaza
|x| this function is known as modulus function or absolute value function
exotic gale
oh ok i thought you were talking about like modulo or something
sharp smelt
I'm working on part (c) I was thinking of the following :
$P=\left{0, 1- \frac{1}{10^{\ceil{\varepsilon}+1}},1\right}$
west sapphire
what's the complicated 10^ceiling(eps) doing for you in the denominator?
sharp smelt
what's the complicated 10^ceiling(eps) doing for you in the denominator?
basically ensuring no matter how large my eps is It will be less than 1+eps
well, I suppose the eps in the num should be 1
will make stuff much easier
jolly parrotBOT
wai
west sapphire
maybe try setting P = {0, 1-c, 1} and compute U(f, P) for that, then solve for what c works
i get a much simpler result that way
sharp smelt
so this may work but is overcomplicated
west sapphire
yea i think so
sharp smelt
c=eps works
west sapphire
yep i got the same
west sapphire
haha being right is more important than simple, but simple is nice too
pearl pondBOT
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