#help-33
1 messages · Page 232 of 1
wai
The intersection of two subgroups is a subgroup..Let $x \in g(A \cap B)$. The $\exists t \in A \cap B: x= gt$. But as $t \in A,B$ which are both normal,\exists t' \in A\cap B : gt=t'g$. So $x \in (A \cap B)g$. So $g(A \cap B) \subseteq (A \cap B)g$. $(A \cap B) g \subseteq g(A \cap B)$ similarly. So $g(A \cap B) = (A \cap B)g$
A,B being normal does not mean gt=tg
wai
The intersection of two subgroups is a subgroup..Let $x \in g(A \cap B)$. The $\exists t \in A \cap B: x= gt$. But as $t \in A,B$ which are both normal,\exists t' \in A\cap B : gt=t'g$. So $x \in (A \cap B)g$. So $g(A \cap B) \subseteq (A \cap B)g$. $(A \cap B) g \subseteq g(A \cap B)$ similarly. So $g(A \cap B) = (A \cap B)g$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.1422 ...t \in A,B$ which are both normal,\exists
t' \in A\cap B : gt=t'g$....
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./796939110815629322.aux)```
thats also not what it means
yes
so how is what I'm saying wrong
sure
so I have to get around that
I;ll work on this
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✅ Original question: #help-33 message
$t'g=t''g$ with both in $G$. Then $t'(gg^{-1}) = t''(gg^{-1}) \implies t'=t''$
wai
does this work
yes
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what is the difference between the two formats
i had a different ticket opened before and i thought i was crazy but in that ticket i used 0 as true while looking at the left table
if i wrote our the table on the right in using both ttff format and 0011 format would they be Logically equivalent
By using TF's and 01's you mean?
Nothing really
You might get the TF
In formal logic
And 01 in Boolean Algebra
But they are essentially the same
i got myself mixed between the two and stared interpreting 0011 as ttff
Yeah, no, 0011 would be fftt
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Integration of sin inverse x whole square
,rcw
$\int\arcsin^2(x)\di x$
𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³
if i was your teacher i would ban you from you pens
what kind of typing is this
even hebrew is more understandable than this
Dude, if you have nothing to contribute to the problem, don’t say it.
Exactly, so don’t even talk about the penmanship.
I wrote it in latex above you.
still, it’s irrelevant to the problem. No need for the comments.
whatever makes you sleep at night
!nosols
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
what is not understandable?
the typing
bro i think its just you
delete the answer that you just sent
it doesn't matter what the argument is because it's still in respect to the argument inside the sine
and everything else is understandable perfectly
I also typed the prob bro 😭
And my handwriting isn't that bad 😭
Arcsin ?
means the same thing as inverse sine
i think its a combination of IBP and trig sub
Ohh ok
But it's wrong
try integrating by parts and then using sine substitution
$t=\arcsin(x)\Rightarrow x=\sin(t),dx=\cos(t)\di t$
^
wdym it’s wrong
𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³
Uhh acc to the awnser key
arcsin = sin⁻¹
sin^-1 is csc
it is
It is not.
sin^(-1)(x) ≠ (sin(x))^(-1)
no it doesnt
it is. Read alex’s answer.
am out
This is the ans
Yes.
Try applying IBP first
And then you'll probably get some trig substition there
u = arcsin²(x) and dv = dx
try from there
you can also go from there right away if you want
Ok just a sec
but i'd go like this
after doing all this you would do int by parts
-# talk more infinium
-# the more messages the faster you become green 
-# im still pissed after that sekiro came in here to talk about penmanship instead of actually helping 
Wait I think I solved it
Show your steps!
Few more mins
okie dokie.
Show the work.
good answer
👍😃
Yoo no way
U actually read the manga
"We are gonna study ethics" or whatever
I forgot the name
It's very underrated
!done
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do you have a question?
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cool
No
No
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I have this question with my math work that I've been struggling with-- I understand how to do part a, I just don't understand how to translate the answers from that into the table in part b
maybe it would help if you can share the answers from part a.
You need to find interest, principal and new balance for 3 months
here’s what I put in the calculator
you have three screenshots here for two blanks in part a though.
the last 2 screenshots was me calculating the interest the way my professor told me to
Why am i getting 582. 35
so screenshot 1 is the monthly payment and ss2-3 are the total interest
Yeah but ig she’s done some calculation mistake if im not wrong
I can't tell what I typed wrong honestly i keep retyping it and i dont know if my calculator is just a bad one or
No ig, cuz if you did it by calculator then you might he right
@faint pewter Has your question been resolved?
@faint pewter you gotta find this for b part
@faint pewter Has your question been resolved?
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can anyone help with me question a) and question b)
the marking key for question a) says indepdent events
but if its a bernouli, why are we concerned with independent events
isnt that concern if we are considering if it can be binomial
heres the markig key
its very ambiguous
the probability of success has to be the same every time the experiment is conducted
wait a second i dont understand this
isnt bernouli concerned with only 1 trial
why is it important for probability of success to be same for every time the experiment is conducted
if it changes, cant we define new bernouli for each for those indiviidual trials
I'm not sure what is confusing you here. A binomial distribution is a sequence of bernoulli trials with the same probability of success
It sounds like you are overthinking the meaning of part (a)
but the question is asking why it can be considered a bernouli trial
and saying that the events are independent sounds like its justifying why it can be considered as binomial instead of bernouli
ye i think im overthinking but why must events be independent for it to be able to be considered as a bernouli trial
I don't know where you are reading independence in part (a)
the answer key says that "indepedent event" is a reason why it can be considered a bernouli trial
but i dont understand this marking key because it seems to be justifying why it can be considered as binmial instead
If you have two independent coinflips, you can think of it as two bernoulli trials or one binomial depending on your perspective
this is the marking key
oo yea thats right
but why must it be indepedent events, like non constant probability of success
cant we define new bernouli parameters for each trial
If you have two dependent coinflips there are 4 outcomes not 2
the marginal distribution is bernoulli but the joint is not
how is there 4 outcomes
isnt there only head or tail outcomes
but the probaiblity are different
I said for two dependent coinflips. There are HT TH HH TT
At the end of the day it depends on what your random variable is tracking
but this is binomial right
we are no longer tracking 1 trial
You asked why can't they be dependent events
ye they cant be depedent for binomial distrbutions, but how about bernouli
bernouli context dont even consider whether there are other trials or not, arent bernouli only focused on 1 trial
wait its okay i think its the question, how thye phrased it
they said why this can be considered a bernouli trial
but i still dont understand onn how i should solve question b i)
and the marking key also dont make sense
the marking key does this and i dont understand why do they this
and why do they not include the P(X=0) when considering P(Y=3) and missing P(Y=0) for P(X=3) case
this is question b i)
cann anyone help
@tight lance
what do you need
i dont undestand why the question do P(X=0)+......P(Y=0) as a method to solve question b i)
and a big issue is that i dont truly understand what question b i) is actually asking
anyone helpers online?
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Is my proof fine?
"only if" part is fine
"if" i am not quite sure what you are saying, what do you mean by smallest positive real number?
also the negation of < is $\geq$, just fyi
Aude
I use proof by contradiction
Epsilon not defined to be the smallest positive number?
ah yep
what is the smallest positive number
So |a-b| > epsilon
can you find it?
Wdym?
Epsilon
Is the smallest number
Like the definition
Nope
there is no smallest number
What
epsilon by itself is not defined to be anything it's just a placeholder for a set of values which are defined in the problem
you will learn about this when you continue with your study
i am pretty sure this exercise is in abbott
So how to proof it
i won't tell you how but i will help guide you
For starters, you wrote you very first line incorrectly.
you are correct to use contradiction though we just need to clarify some concepts
if epsilon is the smallest positive number then what is epsilon/2?
I think the problem in forall not?
We need to find some value
It make forall wrong
Ah that make sense
What you wanted to write was
$$\forall (a,b\in\bR)(a=b\Leftrightarrow \forall(\varepsilon\in\bR^+)(\abs{a-b}<\varepsilon))$$
SWR
But why i see in books that epsilon is placeholder for ver small value?
that is its typical use
in analysis
but that doesn't mean it's the smallest value
well, because one doesn't exist
It can be arbitrarily small, but there is no smallest value
think of a and b as two points on the real number line and |a - b| as the distance from a to b
what this definition is saying is that if this distance is arbitrarily small, as SWR said, i.e. for any positive distance you could possibly imagine - think of any insanely small number
then these two are actually equal
But that mean a-b=0
If there insanely small number
Is there a number smallest than it not equal to 0
Let 0.00000000001
Nope
There 0.000000000000000001 smallest than it
it's saying that for any positive distance someone throws at you (which we will define as epsilon), the actual distance is less than that
and we know that |x| is always 0 or more
so if it can't be more
then it must be 0
but |a - b| is zero IF AND ONLY IF a = b
which you can and also should prove
does this make sense to you?
Some properties of $\bR$: if $0<x$, then $0<x/2<x$. That is, no matter how small a positive number is, there is always a smaller one
SWR
From that i conclude that the problem is forall epsilon
yes, that is in the problem statement
If the statement is
There exist epsilon
Again, your problem statement should be written like this
Then a=b not necessary
a = b is sufficient (but not necessary!) for "there exists eps in R^+ such that |a - b| < eps"
Can you use that bot to organize it
sure thing
$a = b$ is sufficient (not necessary) for the statement $\exists \epsilon \in \mathbb{R}^+$ and $|a - b| < \epsilon$. For two statements, $A$ is sufficient for $B$ is logically equivalent to $A \implies B$. So $a = b \implies \exists \epsilon \in \mathbb{R}^+$ such that $|a - b| < \epsilon$
Aude
Thank you
we use counter-examples to disprove statements
your instinct to use contradiction is actually the most elegant approach
and it is actually the one thing that ties this problem to the "for all" quantifier for positive epsilon
What you think? @still temple
First looks good. Reading second
Looks good. Just one question: what is your proof that $x\ne0\to\abs{x}>0$? (That is, is this something you have proven already?)
SWR
hello! Sorry! Was making tea
you can take SWR’s word though if you are in a rush, I am sure he is more qualified than me
Good!
Ok
Also make sure you understand why you can let eps equal the distance between a and b
It’s precisely because of what I said before
This statement holds for all real positive numbers
So that’s why we can choose epsilon to be |a-b|
And we don’t even know what it is - just by the virtue of it being non negative and nonzero it is automatically a “candidate” for epsilon
And if you are done, feel free to type ,close @granite dove
I went to eat and now I'm back. I'll solve what SWR sent.
Cuz the statement said for all epsilon
And |a-b| > 0 so epsilon can be |a-b|
For all a,b>0 ,There exist epsilon>0 such that epsilon=|a-b|
Maybe the quantifiers applying is ass but i understand the concept
@proud ice what you think
yes!
Can you see that for me? @still temple
Looks fine. Proof might be a little long but still fine
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I see what they did but idk why they did it
How did they decide that r = 3 and the number of elements in X = n
and the answer was 56
Normally these questions involve slots and things that go in those slots
I cant tell what are supposed to be the slots and what are supposed to be the things in the slots
if that makes sense
the number of subsets of size k of a set of size n is (n choose k)
you choose the elements in the set
Julian
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So the subsets are the slots?
And the elements are the things that go inside?
sure 3 slots
the subset itself is a slot
But how can an element be a slot sry
it goes in the slot
if we have n marbles in X
and 3 slots
its saying 56 different ways to fill the slots
but order doesnt matter
Ok just so that I understand, n choose k, n is always the number of slots and k is always the things i need to pick, right?
just generally not specifically for this question
n is the size of X, k is number of slots, and n choose k gives us the amount of unordered ways to fill the slots
for example if we have the set {1,2,3}
n=3
and we want a subset of size 2
so 2 slots
we have {1,2}, {2,3}, {1,3} = 3 total subsets or ways to fill the slots
we dont count {2,1}, {3,2}, {3,1} cuz order doesnt matter
Ohhh
its kinda confusing the first time maybe watch a video with examples
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why does this limit equal 1/2?
assuming it actually does
intuitevly n^2 - n is approximatly n^2 where n is large
and so your expression is approximately n/(sqrt(n^2) + n) = 1/2
to do it more formal, divide both numerator and denominator by n
and then this should simplify nicely to an easy limit
but how do i divide the denominator by n when it has that root
oh like the root divided by n + n/n?
so
sqrt(n^2-n)/n = sqrt(n^2-n)/sqrt(n^2)
=sqrt((n^2-n)/n^2)=sqrt(1-1/n)
and 1/n tends to 0
so its sqrt (1-0)
=1
plus the other 1 we have 2 in the denominator
1 in numerator
oh so you made n sqrt(n^2) to work with the root
and the division by n is like a standard thing when calculating limits
i think i understand it now
thank you
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✅ Original question: #help-33 message
wait sorry i dont get this step
or uhhh 1-1/n?
I can try to help also
im on this step rn
Okay so you have an algebraic\manipulative method
and you have on the other hand L'Hopital's rule
L'Hopital's rule is applied once you learn derivatives
But, if that isn't the case for now, we resort to algebraic manipulation
So here, multiply by the conjugate of the denominator
rationalize the denominator
So, it becomes:
$\frac{n\sqrt{n^{2}-n}}{n^2-n+n}$
wait . . .
So, it becomes:
$\frac{n(\sqrt{n^{2}-n}-n)}{n^2-n+n}$
and all of this leaves you with something ugly
My math isn't mathing, nvm
what I suggested at the beginning is good
Perhaps factor out n^2 from the radicand.
$\sqrt{\frac{n^2 - n}{n^2}} = \sqrt{1-\frac{1}{n}}$
ExpertEsquieESQUIE
is this what you don't understand?
i think i missed the mention of how 1/n is going to 0
so its 1-0 under the root right?
yes
yeah sorry im scatterbrained
all good
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i did a proof by case here. to show if x is rational, all the following statements are also rational. however when it comes to x being irrational, idk how to prove (ii) and (iii) to be irrational too
i mean seems like you've already done it then?
you'll need to show four things
- if (i) is true, then (ii) is true
- if (ii) is true, then (i) is true
- if (i) is true, then (iii) is true
- if (iii) is true, then (i) is true
(technically, you don’t need to show four, only three, but four is probably easier here)
yeah, technically 3
- if (i) is true, then (ii) is true
- if (ii) is true, then (iii) is true
- if (iii) is true, then (i) is true
basically, you need to set up some chain of implications such that any statemeny implies any other statement.
oh yeah i can just do a contrapositive of a biconditional oh... normally for p <--> q we have to prove p-->q and q-->p, here we just prove neg p-->neg q and reverse?
okay yeah i forgot i can just do that ty
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Okay so how he got a-x after observing
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Which version do y'all prefer?
The underlined one feels a little redundant, but also kinda sounds better.
"$\in \mathbb{G}$" and "is a member of $\mathbb{G}$" are pretty much the same. both are accurate and widely used \
personally, though, i think you may want to go with "$a,b,c$ be members of $\mathbb{G}$" if you'd like to keep that text
ηασιβ ♥
To understand what you mean: Are you saying, either keep the text or remove it, but don't have both? like, don't have " in G be members of G" . Or are saying that its fine in any combination?
it is fine in any combination, because it's understood what the elements are, and what the bigger set is. that's really all that matters.
however, "in G be members of G" is technically redundant, so if we're being really specific, i would go for only one of them
my personal process is this: the "natural" thing to put inside of a group is an element, so if you say $g \in G,$ that's going to be unequivocally understood as an element of a group. if, however, i thought there might be ambiguity, i might say "let $(gh)^2j^{-1}$ be an element of $G$", so it's clear what $\in$ is doing here
ηασιβ ♥
I initially had the first one, just "Let 𝔾 be a group and 𝖺, 𝖻, 𝖼 ∈𝔾.", but it almost sounds like the sentence cuts off, when I read it in my mind. But, if it seems ok to have that, then I'm gonna go for this version.
I was just concerned that it might have been "wrong" in some type of way.
Thanks for the feedback. I appreciate it.
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i mean seems like you've already done it then?
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Q3
This is under chapter differentiability so ig its gonna use that topic
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Let $f(x)=x 2^y + y2^{-x}-x-y$ and note that $f(0)=0$ (which means the two sides are equal when $x=0$)
Civil Service Pigeon
you should see from this that ||it suffices to show f is never zero again||
Further hint: ||is f decreasing? increasing?||
@hazy fox Has your question been resolved?
Yes so how do i show this part
Couldn't you take the derivative to show if it is decreaing or increasing, and then from that show that f will never equal zero again, and hence the two sides will not be equal?
Hello
I am taking Differential Equations next semester, but that is just my guess
Est ce que une application peut etre non injective ni surjective ou bijective
?
Donner moi un reponse
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@marsh citrus est ce que une application peut etre non injective ni sujective ou bijective
Its a multivariable function
Well yeah but take the partial wrt to y and divide it by the partial wrt to x
I thought that works
Idk how to find increasing decreasing nature of a multivariable function 😭
Its not in our syllabus either
Yeah I am pretty sure that derivative dy/dx is just partial of y divided by the partial with respect to x, then you can check if it’s increasing
Try that
That’s just my guess
Oooh is that so
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how to solve (X+3)^3 + 64?
i mean how to factorize it
Look at it. Is it in the form a^3 + b^3?
Do you know how to factorize a^3 + b^3?
yea
What is it?
(a+b)(a^2-ab+b^2)
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Need help understand the expression I marked up in the red box
How does subtracting two points and dividing them by their distance=x²? (Aka 3²)
A(3.001) is the area from 0 all the way to the second of those two red lines
and A(3) is the area up to the first of them
so A(3.001)-A(3) is the small area between them
the one in yellow
this area is approximately a rectangle
with height 3^2 and base 0.001
so if you divide its area by 0.001 you get 3^2
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The data A has 4 elements and the average is 5,the standard 편차 (표준편차) is 2
And the data B has 6 elements and the average is 5, the standard편차(표준편차) is 3
Then what is the standard편차 of both sets of data A and B?
Since both their mean is 5, so combined mean will be same right?
What the heck?
<@&268886789983436800>
Irrelevant bro delete
Maybe not if it has something lower than 0?
You can apply formula for combined variance when means are equal
No they said average is 5 for both right so mean will be 5 for both
I have no idea whether the average of combination of both datas will still be 5 or not if it is a variable
We compare overall mean no matter what their individual values
Its in the question right?
The average of combination of data A and B
It will be same
average of both datas will be $$\frac{\sum n_i x_i}{\sum n_i}$$ , x_i being the averages and n_i being the number of averages of each dataset
RadMeerkat62445
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
so 5
Take a data set of 4 numbers (2,4,5,9) and another data set of 5 numbers (3,4,5,6,7) now take their averages individually and take their combined average
It’ll be the same
There is a chance of either data A or B containing something lower than. 0
Do it with (-2,6,7,9) and (-5,0,8,10,12)
It’ll be the same mate
When it says average is 5, then it is 5 no matter what numbers you use
Calculate individual and combined average for both
I think it was true
Maybe the thing that you said
Yeah, remember when it says average is 5, then it is, no matter what number combinations you use
So now we have combined mean 5 right
So use combined variance formula for combined mean
Let us say we have a dataset of m elements $(a_1, a_2, .. a_m)$ and of n elements $(b_1, b_2, .., b_n)$.
Then, the average of the dataset of m elements can be denoted A and is given by $\frac{a_1+a_2+...+a_m}{m} $ and that of the dataset with n elements can be denoted B and is given by $\frac{b_1+b_2+...+b_n}{n}$. But now $mA = a_1+a_2+...+a_m$ and $nB = b_1+b_2+...+b_n$.
We want to find the average of the combined dataset $(a_1, b_1, a_2, b_2, ... a_m, ... b_n)$.
We remember that the average is the sum of all observations (i.e., the sum of all elements $(a_1, b_1, a_2, b_2, ... a_m, ... b_n)$ divided by the number of observations. In our case, we have m observations from the first dataset and n observations from the second dataset, so we have m+n observations.
Then, the combined average C is $\frac{a_1+b+1+a_2+b_2+...+a_m+...+b_n}{m+n}$. But from above, $mA = a_1+a_2+...+a_m$ and $nB = b_1+b_2+...+b_n$. Hence C = $\frac{mA+nB}{m+n}$.
RadMeerkat62445
@faint grove Has your question been resolved?
@faint grove You dont know the formula for combined variance?
I don't know
Alright, write it down then
@faint grove You get it now?
Lmk if you want me to explain it with our question
Yes
We have a data set of 4 elements
So n1 = 4
With standard deviation 2
So square of standard deviation is variance
So sigma1^2 =2^2
Can you try for n2 and sigma2^2? @faint grove
( 4 * 2 * 2 + 6 * 3 * 3 ) / 10 then?
Yeah
Then what you’ll get is variance
And sqrt of variance is standard deviation
Its done
@faint grove Got it?
Yes
Lmk your answer so can verify
7
SqrtVariance = Standard deviation
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Question 9
My thought process yet
Use rolles theorem so there exists a c for which f'(c) and f''(c)=0 for some c€(a,b)
Trying to define a function on which i can use imvt but i cant put my hands on that part
you are close
the two c's that you get from rolle's theorem are not the same necessarily
and yes, try to think what can you use IVT on
also I think your function needs to be continuously second differentiable
Yep
Yes i cant think of it 😭
the function in the equation 
if i consider that as a function i would get f'''
I dont have any info about f'''
the function in the equation is
f'' - lambda (f')^2
Yes but rolles theorem is f'(c)= 0
yes
The derivstive of that equation will give me f'''
@hazy fox Has your question been resolved?
lol
hey
bro
this is NOT highschool math
@hazy fox
dont we have to use
rolles theorem
@hazy fox Has your question been resolved?
Its entrance exam prep for college
So technically pre uni :😭
Yes
You're not supposed to take the derivative
We don't know is f''' even exists
What is the function i should assume then
/what to do
@cunning fiber help him out bro😪
you'll want to use IVT here
Yes i figured that much out
so you need to find a point where f''+ alpha f'^2 >=0 and another where it is <= 0
for that you can look at points where f' = 0
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Need help with quadratic equations
send your question(s) here
the vertex of a parabola is its "turning point" i.e. where it switches from going up to going down (or vice versa)
looking at this graph can you scribble over it to pinpoint where the vertex is
(or if you feel up for it, name the coords right away -- but a visual will do fine)
What are its coordinates?
I honestly do not know I don’t know what I’m supposed to be looking at
right, would've liked a more precise reading
tho you are in the right ish place
gimme a sec
This point
You see it's derivative there is 0
this is way way too early to be talking about derivatives
OP, maybe it helps to think of the vertex of a parabola as the "tip" of the parabola.
I see that’s 8 right
That's not the point of what I'm talking about
Yes
8 is its y-coordinate
So what is the tuple
i would bet €20 op does not yet know what a derivative is
given he is just studying quadratics atp
not sure if we should also use the word "tuple" here too, given the same thing.
Wouldn’t X be -3
also yes i would ask for coords
the point is sitting directly on the y-axis
what should its x coord be
(also dont confuse this point with other pts of interest on the parabola)
So what should I be looking at for the X cord
@umbral oasis i think my point is quite clear, your explanation is overengineered to the point of not connecting with op's knowledge. simple as that.
what is the x-coord of a point that's sitting directly on the y-axis?
this doesnt require you to know anything about parabolas specifically
just reading coordinates from an xy grid
It's not overengineered. I literally used the word derivative once. If he doesn't understand it he can choose to completely ignore that part of my "explanation"
Or ignore my explanation entirely
10?
no
Idk what I’m supposed to be looking at to find the x cord
you're not supposed to be like, "finding" the x-coord by means of a formula
are you OK with us going over some basics of coordinates real quick
Yes
ok right
the graph is not fully marked
can you see the y-coord of the vertex now
we're trying to establish what the x coord is.
oh okk
We already have y we need X
you know the point where the x and y axes meet?
this point has a special name
do you knwow this na-
augh.
well, i can't really continue my explanation if other people are going to talk over me, i think.
@foggy sphinx the point where the x and y axes cross each other is special enough to have its own name. do you know what it is called?
No
A lil
does it ring enough of a bell that you can tell me what the coordinates of the origin are?
(again, forget about parabolas, we're just going over coordinate grid basics)
I feel like it’s zero
a point on the plane has two coordinates, written in brackets (obligatory, you can't drop them, ever) and with a comma in between.
Huh?
it takes two numbers to tell where a point is
an x and a y
they are written as follows: (x,y)
i was also emphasizing that dropping the brackets and writing x,y is wrong.
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/cc-fifth-grade-math/imp-geometry-3/imp-intro-to-the-coordinate-plane/v/introduction-to-the-coordinate-plane
Introduction to the coordinate plane. Plotting points on the coordinate plane.
go give this a watch tbh
K
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@foggy sphinx Has your question been resolved?
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looks like a calligraphic C, $\mathcal{C}$
κλαουντ ☁ (cloud)
they don't look exactly the same because they are in different fonts (the C in that message appears to be from the ams euler calligraphic font)
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please help me solve. i don't even know how to start solving this
do you know how to find intersections of straight lines on a coordinate plane?
no.
🙁
What topic are you doing right now?
this is the bonus problem of my alg homework
actually wait let's go further back
if they asked for the length of AB, would you know how to find that
ok
58?
What was the homework about in particular?
But what if F isn't perfectly on the point
this is a square grid
- i am tracing a horizontal thru C
even i know that
<@&268886789983436800>
WHAT
<@&268886789983436800> SPAMMER
Yikes. Has been muted and presumably banned
yuck
did someone @ me?
mistaken modping
lol
might wanna change your nick
i'm very afraid rn
calm down, it's been handled already
i was looking for help with math...
ok
I was thinking about setting axis and finding the equation of the lines ab and dc and then find the point e as their intersection
i tried feeling out that route
but op said he doesn't know how to proceed from that
this could also work if u know what i mean by that
i can do that
if you know how to find line equations then finding the intersection is a very simple idea
can u notice sth with this pic
thats not really true
also they are all parallel
what's the proportion of (red+orange) : yellow
1:1
sorry, (red+orange) : yellow
2:1
yeah, and can you find the total length of the colored segment?
i.e. red+orange+yellow?
1+sqrt(2)
hmm not really
how did u get that?
notice that the endpoints lie on the grid
so u can just pythagoras it
yes
i think u did ur pythagoras wrongly then
can u now find the distance from the left endpoint of red to E?
np
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Linear transforms
What's your question / what have you tried?
!status
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
.close
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Sorry for the amount of questions I’m asking but I’m trying to solve for #17 b.
and im at the point where im finding the limits
based on what ive done for the cusp im tryna do the same to see if theres a vertical tan line but
would that mean i would have to find the limits for all 3 critical points
or is there some shortcut i can take/ are there steps that im doing wrong
Is 6 a critical point? Is it not just 0 and 12?
But anyway, since f'(6) = 0, we can tell that it's going to be flat rather than a cusp or vertical tan
Well the other two have f'(c) approaching +- infinity, which implies a cusp or vertical tan
A cusp is what you have in (a) right? Where the function makes a sort of spike shape?
yeah i didnt solve it on the paper but its a cusp on a
i was doing it in my notebook, im still a little iffy on it but i do get it
Yeah, so if you remember how the way a curve tilts up/down relates to the positiveness/negativeness of the derivative
You can see how the critical point in (a) is a cusp because f' goes to negative infinity from the left (So it's going down), while it goes to positive infinity from the right (so it's going up)
While if you had a vertical tan, it would have to be approaching either negative or positive infinity from both sides
So once you've found either a cusp or vertical tan, you can tell the difference by analyzing the sign of the derivative
Does that make sense at all?
could you explain this part a bit more
Sure thing
How about looking at this graph of your work from (a)
The blue line is your original function, while the green one is the derivative
You can see that the derivative is going to -infinity from the left, so the curve is going down really fast
oh my god this visualization actually just simplified it so much
😭 thank you anyways yes continue
While it's going to +infinity from the right, so it's going up really fast!
Lmao no problem, graphing a problem is often a great idea if you're having trouble understanding it :)
So a cusp is when the derivative is going to +infinity from one side and -infinity from the other, causing that sort of spike shape
While with a tangent line the derivative will go to one of -infinity or +infinity from both sides, since the curve won't be changing direction as it goes through the critical point
Remember, positive derivative means it's going up, negative derivative means it's going down
that'll be it thank you
Cool, can you use .close to free the channel then?
.close
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How do I solve question (ii)
use the slope and y-intercept in the y-intercept form and you'll have your equation
yes
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😋
Given that they just joined and that username, I feel confident saying they're a bot
.close I don't think there is any real question here.
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an is absolutely convergent and bn is convergent series then their multiplication anbn will be convergent?
I am looking for examples so that i can be sure
What kind of examples are you looking for? Do you have examples for absolutely convergent series?
Ann
the zero series is in fact convergent, yes
I was looking for such examples where it fails if does
you do some manipulations and at some point you say, "it would be nice if b_n were a ____ sequence"
at some point, |anbn| would be smaller than |an|.
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Hey?
hello
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Let ${\delta_1, \delta_2, ..., \delta_k}$ be the set of determinants formed by the set of 9 real and non-zero numbers ${a_1, a_2, a_3, ..., a_9}$.\\
what is the value of k?
What is a determinate for a set of numbers?
nah its okay
just consider if 9!/18 would be correct
im not sure what the answer is
We can't help you with the question if you don't clarify what the question says
i think what's meant is like, all the possible values you can get by sticking the a's into a 3×3 determinant in every which way?
oh i didnt realize he wanted clarification on the question
mb
determinant with the you know
so can the question be rephrased like
How many determinants can be formed by the set of 9 real non zero numbers
sure
uhh wait nvm distinct is not given
but how would you do it if it was distinct
if its distinct it should be 9! right
if its not distinct
that makes it a little tougher
i dont know
cause there are too many combinations
No cuz you can rearrange rows and columns to get the same determinant for different matrices
@sacred idol Has your question been resolved?
@sacred idol Has your question been resolved?
Hmm... what if we look at 2x2 matrices first? So just {a1,a2,a3,a4} ?
Or rather {a,b,c,d}
I think the maximum number of determinants made from 4 values would be 6, from
ab-cd
ab-dc
ac-bd
ac-db
ad-bc
ad-cb
Or... actually 12 because we would need the negatives of all of those
Do negatives count though
duh??
Well 1 =/= -1 so I assume so
Since we chose a1... a9 and not -a1... -a9
Actually it doesn't matter since we can get the opposite determinant by swapping columns
"... by swapping columns" that's what I meant by the negatives
Aren't some of them duplicate?
No! This has all the duplicates taken out
I just also took out the negatives by accident
Or....... No wait I'm dumb????
Too much screwy algebra, I forgot cd=dc lol
Why isn't ad-cd equal to ad-dc
ab - cd
ab - dc
bruh
i would say
Anyway, so take out my doubled half and add in the negatives and we get back to 6
for a given matrix
either a transpose is performed
or/and one of these two choices
- swap rows AND columns
- do nothing
which gives 2 x 2 = 4 matrices in a group with same det
so 24/4 = 6
Well there's only 1 way to do a row + column swap isn't there?
I think these 4 matrices should have the same det
[a b]
[c d]
[d c]
[b a]
[a c]
[b d]
[d b]
[c a]
yeah so thats one choice
and the other one is to do nothing
Those last 2 matrices can't be gotten from the first 2 by row/column operations
Transpose is the other option
So for the 3x3 case we would have at most 9! possible configuration divided by 433 : even number of row operation and column operation, odd number of row and column operations, and twice that number for transpose?
So if it's transpose and double operations... we have 2 * |S_3|^2/2 = 36 redundant permutations?
I got the same
Hell yeah
guess so
But that's assuming we've got them all
ill try with an example
All the ways of making a redundant permutation I mean
we should get 10080
That's a lot of cases to check
I think we can choose 9 transcendent numbers and that will make sure that we can't get the same det with different formulas
Actually square roots of prime numbers should work
@sacred idol Has your question been resolved?
@sacred idol Has your question been resolved?
Wdym
@sacred idol Has your question been resolved?
uhh nvm
