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12:02 AM
people do not seem to have liked this one:
@WillJagy Hello. I am bad at counting.
12:14 AM
@WillJagy How are you?
12:29 AM
does anybody know a good book for functional analysis with a lot of problems
1:07 AM
@pourjour You're taking a functional analysis course?
1:33 AM
$H \leq G$ s.t. $H$ is not normal. Show there exists cosets $aH$ and $bH$ such that $(aH)(bH)$ is not a coset:

Let $b = a^{-1}$. Since $H$ is not normal, not all products of $aHa^{-1}$ are in $H$. By definition of coset, $xH$ is a coset for all $x \in G$. So $(aH)(bH)$ is not necessarily a coset.
Is this correct?
1:56 AM
@AlanH If $H$ is not normal, there exists $g\in G$ and $h\in H$ such that $ghg^{-1} \notin H$. You ought to work with that.
In what you write, what is $b$? What is $a^{-1}$?
@PeterTamaroff I meant to write $aHa^{-1}H$ somewhere in there. $b$ is the inverse of $a$
uh okay let me think about it again
@PeterTamaroff It's neat, but on the border of obvious. It takes a minute to figure out what is going on.
@robjohn Heh, true.
@robjohn Now I have to prove every natural $n$ can be written in $2^{n-1}-1$ ways as a sum of positive naturals less than $n$.
2:12 AM
The basic idea is $(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})\dots=\dfrac1{1-x}$
@PeterTamaroff Case it on the number of $0$s on the right in binary
no $0$s on the right, there are just as many ways as $n-1$
@robjohn Come again?
Sorry, I was thinking of something else with a similar solution. Case on the smallest number in the sum...
@robjohn "Case on..."?
Break up the sums based on the smallest number in the sum and equate that with the number of ways to make up a previous number.
@PeterTamaroff So if $H$ is not normal, there exists some$ g $and$ h$ such that $ghg^{-1}$ is not in $H$. Then since $(ghg^{-1})H \in (gH)(g^{-1}H)$, so $(ghg^{-1})H$ cannot be a left coset?
2:24 AM
@AlanH Could you $\TeX$ that up?
@PeterTamaroff Nevermind. Don't bother with it
@AlanH Wait. What you want to show is that the equivalence relation fails to be a congruence.
Ah, you call $mH$ a right coset, yes?
@PeterTamaroff I think I got it. Thanks.
@AlanH Good. Byes.
@PeterTamaroff See ya. Thanks for your help again today
3:01 AM
In 2-dimensional geometry, is it correct to say that the perimeter of the convex hull of a set of points is the curve with the minimum perimeter bounding all those points?
What's the name of a curve of minimum area bounding all points in 2-dimensions?
2 hours later…
4:49 AM
Hi all
Hell @Bageer
I need some help
I've need to a formula for achieve to this:
Well you can ask and someone might help.
My question is for calculating place of an object in an computer app
4:55 AM
Looks like someone is swimming in rep
1. Maximum mode = when one of the objects is placed in 4792.7 px
my other objects must place in 2266.7 px
2. Middle mode = when one of the objects is placed in 1826.496 px
my other objects must place in 685.85 px
@Bageer i leave MSE alone for half a day and look what happens.
this was not exactly what i imagined my highest voted answer would look like.
Anyone know if there is some sort of and operation for tags, for example suppose I don't have a problem with looking at questions that have homework or calculus but not both, so I would have in my ignore list something like calculusANDhomework.
@AlexanderGruber Yah, in a way it is sort of sad.
3. Minimum mode = when one of the objects is placed in 219.15 px
my other objects must place in -167.6 px
@Bageer i don't think this exists, but you should ask on meta (maybe a feature request? i'd use that.)
5:00 AM
@AlexanderGruber I guess I will do that.
totally, I must say that when the first object moves, the second object must change its position
I check many formula for find a formula for do that in the coorect method
but I couldn't find a good method
I can't find a similar formula for doing that for changing the second object position
I check for finding a same multiple for doing that
but I can't
I check for doing that with some other formulas
but I can't find anything
5:50 AM
Does someone here have Mathematica?
@DominicMichaelis actually, me ... want me to run some code?
@VincentTjeng something like that. This `FindDistributionParameters[
Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
NormalDistribution[\[Mu], \[Sigma]]]` gives an unexpected result and I don't know why :/
 Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
 NormalDistribution[\[Mu], \[Sigma]]]
okay, I run mma 9.0 on windows
Same here
5:53 AM
{[Mu] -> 0.00009999, [Sigma] -> 0.000833768}
The result should be mu -> 2000, and sigma -> 40 in my opinion
but I get the same result as you
I mean the mean of a binomial distribution with n=10000 and p=0.2 is clearly 2000 isn't it ?
yep. reading through the documentation right now
@DominicMichaelis - now I'm confused ... what does this mean?
Table[PDF[NormalDistribution[100, 0.2], x], {x, 0, 200, 1}],
NormalDistribution[\[Mu], \[Sigma]]]
I get {[Mu] -> 0.00992401, [Sigma] -> 0.140346}
Ah I see my input makes no sense. I try to make a distribution over the given propabilities not over the the data
5:59 AM
I see.
I thought I just give the propabilities of the old distribution to estimate another one ...
 RandomVariate[BinomialDistribution[10000, 0.2], 10^5],
 NormalDistribution[\[Mu], \[Sigma]]]
this would be a correct input
Hello, I have an off topic problem I would like to introduce. It is a question I posed yesterday which has not gotten a response.
I do not wish to violate chat etiquette, so stop me if I am.
Here is the link to my problem.
6:19 AM
your problem is not well defined
you sum up to $f(n)$ but $n$ isn't in the Domain of $f$
ah thats a typo, thank you.
@Orangutango this does violate chat ettiquette (see point 3 here), but at least you're being polite about it, so i'm ok with it this time
but is the question otherwise clear? i am searching for an algorithm to maximize the sum.
@Orangutango Does it not maximize if your order smallest^0 +largest + second largest^2+...+smallest^n-1?
@Bageer it does
6:24 AM
Maybe I am just being naive
or not...
The optimal way to sum $\{ \frac{1}{10},\frac{1}{2},\frac{4}{7},\frac{3}{5},\frac{2}{3}\}$ does not follow that algorithm.
I almost had an answer to this question, using a bit of abstract algebra + analysis.
I verified this in Mathematica.
In my post, the last sum is the optimal one.
I can post the Mathematica code to verify if anyone doubts this.
I am honestly surprised
thats why this problem caught my attention! my intuition was totally wrong, and I still don't know exactly why
6:30 AM
That is cool
But post the Mathematica Code :)
was the optimal one the second one you posted
@DominicMichaelis someList = {1/10, 1/2, 4/7, 3/5, 2/3};
Last@SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First]
beat me to it
@VincentTjeng thats a Neat code :)
6:34 AM
and very slickly done
cleaner than my code XD
@Bageer yes, it is
thanks :)
anyway @Orangutango - I think part of the reason why your question didn't receive so much attention is that some of us didn't understand it! :)
well at least I'm not alone then! it's been troubling me for almost a week now.
@MRS1367: hi
well, i'll let you all toy with it as you will. it's very late my time and i've been thinking about this problem for far too long today. thanks for hearing me out, i hope one of you gains some insight on the problem.
7:09 AM
@Orangutango - doing a study for sets of 5 random reals between 0 to 1...
data = Table[someList = RandomReal[{0, 1}, 5];
SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First])[[2]], {i, 1, 10000}];
I got {{{1, 3, 4, 5, 2}, 3605}, {{1, 4, 5, 3, 2}, 2664}, {{1, 4, 3, 5, 2},
452}, {{1, 5, 4, 3, 2}, 659}, {{1, 2, 3, 4, 5},
1834}, {{1, 3, 4, 2, 5}, 450}, {{1, 3, 2, 4, 5},
112}, {{1, 4, 3, 2, 5}, 59}, {{1, 2, 4, 3, 5},
48}, {{1, 3, 5, 4, 2}, 85}, {{1, 2, 3, 5, 4}, 11}, {{1, 2, 4, 5, 3},
21}} for 10000 runs
That doesn't look good :/
strangely, only 12 possible orderings pop out ... not 1, 4, 2, x, x for instance
not sure whether i coded it wrongly or what
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}]]
how can you select 'random reals' between [0,1]
7:14 AM
Whats your definition of random?
Pseudorandom ;)
Is there an easy way to figure out the parity of the permutation?
why do you ask
@VincentTjeng well we can say for sure that the first one is fixed
the smallest number is always the first
oh well... i run 100k times and look what pops out
{{{1, 3, 4, 5, 2}, 35695}, {{1, 4, 5, 3, 2}, 26819}, {{1, 3, 4, 2, 5},
4708}, {{1, 2, 3, 4, 5}, 18862}, {{1, 5, 4, 3, 2},
6257}, {{1, 3, 5, 4, 2}, 959}, {{1, 4, 3, 5, 2},
4388}, {{1, 3, 2, 4, 5}, 1011}, {{1, 4, 3, 2, 5},
623}, {{1, 2, 4, 5, 3}, 213}, {{1, 2, 3, 5, 4},
58}, {{1, 2, 4, 3, 5}, 403}, {{1, 2, 5, 4, 3}, 4}}
@bageer what do you mean by parity? I think it can be done
7:22 AM
The permutation sign (whether the permuations is odd or even)
I have test, almost all the ones you put up and so far they have all been even.
{1,3,4,5,2}-> {1,2,4,5,3}-> {1,2,3,5,4}-> {1,2,3,4,5} is odd isn't it ?
@DominicMichaelis do help me check through my code... I don't want us to be trying to find reasons for wrong results :)
Damn, I have been putting in the cycles into wolfram alpha... its late, okay!
{Signature[#[[1]]], #[[2]]} & /@ Tally[data]
(cycles as in the list)
7:27 AM
{{-1, 35695}, {-1, 26819}, {1, 4708}, {1, 18862}, {1, 6257}, {1,
959}, {1, 4388}, {-1, 1011}, {-1, 623}, {1, 213}, {-1, 58}, {-1,
403}, {-1, 4}}
does anyone find it curious that 99% of the lists have either the 2nd or the 5th smallest as the last term?
Oh no particularly good pattern. I have to say that his problem is surprising me.
no that is expectable
Look at the plot
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}, PlotLegends -> Automatic]]
if your number is very small or very big raising it's power don't hurt the sum
Hi @MaisamHedyelloo
oh, i see.
and by the plot the second smallest is most likely to become the last
which is with our data at over 70 % of the cases true
@VincentTjeng Why do you use two orderungs?
7:42 AM
@DominicMichaelis Ordering@{1, 5, 2, 3, 4} gives you {1, 3, 4, 5, 2}
applying Ordering to that again gives you the original order. a bit difficult to explain but i think its correct from the documentation
I am running for 300 k
in your code you make at first a list of 5 randomreals
Then you make a list of its permutations and on the permutationslist you apply a list of functions namely the sum we are trying to minimize and the identiy
this list you sortby the sum,
but this should give the smallest as the first ?
ah right so the largest is in the end and you take the last one
mm thats why i take the last
quick and dirty code
afterward you take the second part which gives you the permuted list of randomreals
your code seems to work
i mean that it looks correct oto me
8:04 AM
hmm ok
@VincentTjeng I gonna write an answer with our intermediate results
thanks! interested to see the details from a big run :)
2 hours later…
10:07 AM
I posted some things
10:27 AM
@DominicMichaelis nice work. hope it inspires some further thinking from other people
i will try with longer permutations
for longer permutations we need another code
i think we can force the first one to be smallest
not sure wha other simplifications we can make
if you can you should extract some examples for the rare cases like {1,2,5,4,3}
yeah we need to do more analysis before improving the code
if we take your command for a list of length 9 it takes 100 seconds at me
length 10 *
10:39 AM
around there
we could reduce it by faktor 10 with forcing the smallest to be first
in the best case
this sounds totally like a project euler problem to me :D
maybe there is something like that on project euler
perhaps :)
10:58 AM
Greetings all.
@robjohn are you around?
I'm looking for some nice example of limits where $\lim_{n\to\infty} n(b_{n+1}-b_{n})=l>0$, $\lim_{n\to\infty} b_n=b>0$
$\ln(n)=b_{n}$ works for the first one..
@Chris'swisesister I don't think any exist
11:14 AM
@Ethan are you sure?
@Chris'swisesister If $n(b_{n+1}-b_{n})$ is bounded
Then $b_{n+1}-b_{n}$ over $\frac{1}{n}$ is bounded
so that $\frac{c_1}{n}<b_{n+1}-b_{n}<\frac{c_2}{n}$
Sum both sides from $n=1$ to $n=m-1$
For sufficiently large $n$
Sense $\sum_{n=1}^m\frac{1}{n}=\ln(n)+O(1)$
That seems right.
Assuming $b_1$ is defined and bounded, if its not we can always start the sum at a different number and use the same argument
in which case regaurdless we get $c_1\ln(n)<b_{n}<c_2\ln(n)$
for constants fixed constants $c_1$, $c_2$, and for large enough $n$
and it follows $\lim_{n\to\infty}b_n$ is not bounded
Though if we lift the second constraint
The only $b_{n}'s$ that work for the first one, are non zero multiples of the natural logarithm I think
I felt it's something wrong with this question ...
11:31 AM
@Chris'swisesister Pardon me asking, but what motivated you to look for such limits?
@skullpatrol I was doing some research on limits.
12:30 PM
@Chris'swisesister yes
1:11 PM
Hello, if we throw a dice 3 times what is the probability that the first is an odd number if we know that we got 1 odd number?
@robjohn it was about the problem I discussed with Ethan. Now it's clear that point.
1:46 PM
Hi, I have a question. What would I classify a 2D surface that is closed but has a hole in it?
like 2 concentric circles defining the boundaries of a closed surface
torus ?
annulus ?
a torus can be 2D?
the surface of a torus is 2 dimensional
In mathematics, an annulus (the Latin word for "little ring", with plural annuli) is a ring-shaped object, especially a region bounded by two concentric circles. The adjectival form is annular (as in annular eclipse). The open annulus is topologically equivalent to both the open cylinder and the punctured plane. The area of an annulus is the difference in the areas of the larger circle of radius and the smaller one of radius : :A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)\,. The area of an annulus can be obtained from the length of the longest interval that can lie completely inside the ann...
an annulus sounds like the exact definition of the 2nd description I gave
So a torus would be a more generic term for 2D closed surfaces with holes in them?
while annulus is specifically circles?
i missunterstood you
you want the whole thing t obe 2 d not only it's surface
1:49 PM
yes it is just a 2D shape to consider
it has a boundary on the outside everywhere, but basically just has a hole in it anywhere
and the shape can be squigly
and an annulus would fit into this category
2:07 PM
@robjohn: Hello! ;-)
2:18 PM
@PeterTamaroff I want just to start learning them
@Ethan $\log(\color{#C00000}{m})+\gamma+O(1/n)$
@BabakS. hey there! good day?
Good day @robjohn our rescue here. ;-)
@robjohn: I have a question. We know that there is not a general formula ruling the primes numbers. But for many finitely prime numbers $p$ we see that $2p+1=4k+3$ wherein $k$ is some positive integers. What can we do with this case. I am trying to generalize a presentation of a group and in it the above equality is being used frequently again and again intuitively. May I ask you any hint? Thanks
2:38 PM
If $p$ is odd, then $p=2k+1$ for some $k$, so $2p+1=4k+3$. Is that what you mean?
@BabakS. It says nothing but that $p$ is odd.
Ohh yes. Thanks I got where I was mistaking.
@BabakS. misstaking a vampire can be hazardous to your health.
@robjohn: Yes :-( but sometimes I lost the very basic facts. I think I am getting Alzheimer at 41.
Are you math professors/PhD grads?
2:45 PM
@EwokNightmares: I am a lecturer.
I admire math experts so much
3:03 PM
Good morning! :D
3:49 PM
good evening
4:01 PM
1 hour later…
5:13 PM
Anyone here good at scientific notation?
I've started the problem [(5600000)(0.000000081)]/[(900)(0.000000028)] and gotten to (5.6x8.1x10^6x10^-8)/9x2.8x10^2x10^-8)... Then to [(5.6X8.1)/(9X2.8)]X10^? And I'm stuck LOL. How would I determine that exponent?
Based on that my answer would be 1.8x10^? So confused on how to get that exponent though. :/
5:37 PM
Hello, can someone help me please :math.stackexchange.com/questions/422940/…
5:47 PM
@amWhy WTF is the problem of the finite groups guy?
@BandeiraGustavo I don't have a clue...but the attitude is surely a turn-off!
Is the following a typo? If $a \equiv b \pmod{m}$, then for some scalar $c>0$, $ac \equiv bc \pmod{mc}$
@AlanH Such is likely. It holds for all suitable scalars $c > 0$.
Hello people.
6:20 PM
@Lord_Farin $\sup$?
@PeterTamaroff $\inf$ :(.
@Lord_Farin Oh, man. I saw that one coming. Why?
@PeterTamaroff Progress on thesis is too slow.
And how are you doing?
@Lord_Farin I'm hungry. And I have a olympic runner of a nose.
6:36 PM
@PeterTamaroff Interesting; since yesterday, such has presented itself to me as well.
I've used my handkerchief as much today as I did all of last week together.
@Lord_Farin Heh. I just answered. I will get to 800 answers this month, I will. =)
@PeterTamaroff These numbers mean nothing of course; yet I felt somehow satisfied for passing 200 a few days ago :).
@Lord_Farin Good!
(Can't upvote by the way; today, I've burnt through my votes exceptionally fast.)
Some days ago I posted something erroneous.
6:48 PM
@PeterTamaroff I suspect several instances of such unfortunate events in my answers list as well.
$$\prod_{k\geqslant 1}(1-q^kz)=\sum_{k\geqslant 0}\frac{q^{}}{q-1}\frac{q^2}{q^2-1}\cdots \frac{q^k}{q^k-1}z^k$$
The first term is (obviously) $=1$.
On the other hand $$\prod_{k\geqslant 1}(1-q^kz)^{-1}=\sum_{k\geqslant 0}q^k (1-q)(1-q^2)\cdots (1-q^k)z^k$$
If I am not remembering wrongly. I lost the page. =(
@anon Do you know an explicit formula for $\mu_n=\text{ the number of } 1 \text{s in the binary expression of } n$?
Because $$\prod_{k\geq 0}(1+qz^{2^k})=\sum_{k\geq 0}q^{\mu_k}z^k$$
well, there is an explicit formula for the $n$th binary digit, and you can add these up over all $n$, but it will not be pretty
as long as you consider Mod(n,2^k) to be a reasonable function anyway
@anon Said sequence is pretty interesting. You can build it as follows. First, observe the first three terms are 1 1 2. Now, insert a 1 at the beginning, add 1 to the rest elements, and join this to the sequence you had. That is 1 1 2 1 2 2 3. Now, do the same, 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4. This builds up the sequence.
So if you know the first $2^k-1$ terms, you know the next $2^k$ terms.
7:11 PM
In particular $\mu_{2^k}=1$ for each $k$.
Yo @Argon wazzup pal?
@Lord_Farin hello
@Vrouvrou Hello there. How are you doing?
i'm good
@Lord_Farin or @anon i need your help please for a derivative
can you help me please ?
@Vrouvrou I'm currently busy, sorry.
7:25 PM
@Vrouvrou What is it?
@Vrouvrou No clue on that one, man.
>_< no one has an idea
@Vrouvrou Well, it is fairly advanced, isn't it?
7:32 PM
yes but
i posted it on matoverflow i hope that someone could help me
@amWhy Just one instance of you happening to make a mistake does not imply that you should refrain from taking the liberty of editing my posts. In 95 out of 100 instances, your edits will correct genuine flaws.
I appreciate your effort of proofreading my answers. :)
@Lord_Farin I do sincerely apologize. Thanks for commenting here. I rarely intervene in others posts (answerers), and doubt I'll need to with your posts...
@amWhy No offence taken. I'm notorious for skimming over "pathological" cases.
I was still thinking "biconditionally" (given my answer), and jumped in too quickly.
@Lord_Farin Yes, you corrected precisely one example of a pathological post, authored by me, yesterday! :-) Thanks.
7:49 PM
@amWhy Hm, I think there was a language barrier at work there. I meant to write "glossing" in place of "skimming".
@Chris'swisesister Greetings :D
@skullpatrol hi :)
@Chris'swisesister How are you?
It pleases me to have found a use for my knowledge about Kripke structures (for intuitionistic prop. and pred. calculus).
@skullpatrol in a bad mood. How about you? :-)
7:56 PM
@Chris'swisesister I'm Ok, thanks :-)
@Chris'swisesister Did working on those limits cause your mood change?
(Two were already deleted.)
@skullpatrol In a way, yes. I'd like to be able to finish calculus problems without pen and paper ... but I still need to work to go there ... (I'm not serious enough in my work - - I need more discipline)
I'm out, going to try to do some useful stuff.
Bye all.
@Lord_Farin hi & bye
8:07 PM
@Chris'swisesister Same. :)
@Chris'swisesister Like any skill it takes time & effort to develop...
@skullpatrol yeah ...
@Chris'swisesister It's like the difference between the ability to do something and the ability to do something well.
@skullpatrol there is also a trick here. Since we're smart beings, it's also possible we try to explain somehow why we're in a bad mood (since we try to explain everything), but it's also possible that sometimes there is about some neurotransmitter imbalances due to excessive work on things, stress.
Hello, all, hello, @skull
8:20 PM
@amWhy hi
@Chris'swisesister :D
Hi @amWhy
@skullpatrol I just get frustrated, that's all...but some things are just part of life I guess.
@skullpatrol :D
@amWhy :D
8:23 PM
Hello :)
How are ya this lovely afternoon?
Fine thanks, how are you?
Well, could be better. Trying to finish up some math homework with the worst textbook ever once again! Thank goodness it's the last assignment LOL.
Orly? who wrote it?
8:26 PM
Do you happen to know if there is a good thread on simplifying exponential expressions with variables only containing positive exponents (assuming all variables represent nonzero real numbers)?
Pearson custom math text- I'd assume our professor picked the content he wanted in the text. sigh
I'm terrible at math other than statistics LOL... And I'm trying to do a math for teachers course :/ Already did a BA majoring in both Psychology and Sociology... Yet this math stumps me biiiig time LOL!
Like if I tried to do 4(5a^4)(2a)^-2 would it just be 40a^2 ?
Hmmm... I would suggest you focus your professors notes.
There is none- it's a distance course. We just have the text. :/
There's nothing with negative exponents in that section LOL
@Shayna Real multiplication is commutive, yes, but 4*(2a)^-2=a^-2.
@Shayna Have you tried the Khan Academy on YouTube?
I did try the Khan Academy. Problem is I just don't really know what I'm looking for most of the time. :/
8:33 PM
Polynomials would be a good start.
@DenisRodman Thank you! In the text it says to just multiply 4X5X2... Has me confused LOL it doesn't even say what to do with the exponents
Try "Rules of Exponents"
@skullpatrol Here's the thing. I can do polynomials. I'm just unsure where to start when it has the 2 sets of brackets and numbers outside of those brackets :/
@skullpatrol Thank you! :D Hopefully I can find something!
Start from the inner-most brackets.
So the 5a^4 and 2a?
If I were to follow FOIL wouldn't it be 4(5a^4) and (2a)^-2?
8:38 PM
What does "FOIL" stand for?
I meant that as a rhetorical question(!)
first inside outside last
but then I wasn't sure if I should be following FOIL or BEDMAS :/
Yes, you need BInomials to multiply to use it :-)
Hahaha ohhh man I feel sorry for the kids I'm going to have to teach math to! I hope I end up teaching like grade 4 or lower!
The FOIL method is for multiplying binomials.

BEDMAS is for order of operations.
So I do bedmas right? LOL
so basically work left to right on 4(5a^4)(2a)^-2 ?
8:43 PM
You need to know the "Rules of Exponents." :-)
Yeah I'm looking through the videos on Khan Academy... I just can't find anything with more than one part to the exponents questions :/ I know how to do the easy ones but when they end up compounded I get confused.
@Shayna Try this to start with...
...math builds on itself.
Yeah, I just watched that one before you linked it. I know how to do questions like those ones. It's just when they get more complicated I get lost unless I have pretty much an exact example to work from. :/
Like... I can do exponents well enough that I got 98% in Statistics LOL
@Shayna 4(5a^4)(2a)^-2 = 5a^2
Okay I see where the a^2 came from... But how did we get 5?
8:56 PM

(ab)^m = (a^m)*(b^m)
00:00 - 21:0021:00 - 00:00

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