#help-13
1 messages · Page 191 of 1
tan30=CB/BB1?
no
then I don’t understand how to apply this in my task
but why is it wrong if this way I find BB1 and this is the height
and in the formula for the area of the base you need the height
Because you didn't show how you got to that part
I have a method of solving it in my head
It likely doesn't allign with how you're doing it
That's why I said it was wrong, but really I meant to say "I don't know"
I need to know your method to understand if we do this
Do you get it?
like yes
so far so good?
hmmmmm.... yes
If you increase the 30 degree angle... the width of the base of the prism would change, right?
So would the height or something, probably
right
Meaning there's a relation between the 30 degree angle, and angle at point c
Find that relationship, and you'll be able to solve the question.
I know that a leg opposite an angle of 30 is equal to half the hypotenuse, but that doesn't fit, does it?
I'm not sure myself to be honest.
Assuming the height is static, which it is
The angle at C is dependent on the angle at B1, probably by some factor of the static height given, which is 2 * sqrt(3)
I'm thinking it might be possible using similar triangles.
At this point, I'm giving ideas lol
what is cubert(3)? 3^2?
ohhhh ok
sure
If AB is double of b
Then would AB1 be double the length of CB? 🤔
I'm not sure... I feel like there's way too little fiven info in the question
Alright, man... I tried.
I'm sorry ;----;

See, I'd spend more time on it, but it's 3 32 am and I have maths of my own to study too. I'm sorry I couldn't help much lol
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Does this look good?
sorry if its hard to read
The quadratic formula:\~\
For some $ax^2+bx+c=0$
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
FancyBredFries
that 2a goes to both -b and the radical
did i mis input
which part?
the first line
oh
i was suppoesd to divide 16 by 2
at the start
so in the end it would actually end up being {-4 +/- square root 79}
?
^
pretty much all ur calculations are wrong 😅
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Factor by grouping 2x^4+3x^3-32x^2-48x
good start! Is there anything out of x^3-16x we can factor out?
right direction, but you're missing a step here
we have to factor out an x
in order to do that
So then you have x(x+4)(x-4)(2x+3)
that's about as far as you can factor
so there's your answer 🙂
Thanks I have a question tho how do we know to factor the x outside the box?
Like how does x^3-16x factor to x
Nvm I got it
.close
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This is part of a chemistry problem, so the ask seems a little weird out of context. I have 3 positive constants $a, b, c$ where $a < b$. I want to prove that $2a < a + b + 2c - \sqrt{(a-b)^2 + 4c^2} < 2b$
Where would I start
I would start by adding both sides of the two inequalities by the square root, then subtracting both sides by 2a or 2b respectively to isolate the square root, and then squaring
After that it's probs just a bunch of expanding and rearranging but I'm not 100% sure
The fact that $a <b$ is useful for the latter portion of the inequality chain cus otherwise after taking those steps, u can't necessarily justify that squaring both sides preserves the inequality
992qqoloy
Sorry I'm a little confused on the notation. Can you write this out
bleh I wanted to avoid that cus I'm on my phone tbh 
oh nw
When we prove this sort of thing, we need to start with a true statement right?
So what statement do I start with? $a<b$?
Methylfluorophosphonylcholine
Well with inequalities like this
What you generally do is work backwards
That only works if all the steps you do are reversible tho
So for instance if $x^2 < y^2$ you can't generally conclude that $x < y$
992qqoloy
Buuuut if say, you know that $0 < x$ and $0 <y$ then you can.
992qqoloy
So for simplicity's sake imma call the whole expression under and including the square root $\sqrt{n}$ 
992qqoloy
makes sense
U split up those inequalities u want to prove into $2a < a + b +2c - \sqrt{n}$ and $ a + b +2c - \sqrt{n} < 2b$
992qqoloy
I was thinking first, for each of those inequalities, you add $\sqrt{n}$ to both sides
992qqoloy
Then for the first subtract $2a$, and for the second subtract $2b$
992qqoloy
And then square both sides so u can get rid of the square root
The tricky part here is that u gotta make sure $x^2 < y^2$ implies $x <y$, i. E. That the step is reversible
992qqoloy
mmmm
for the first inequality, since $\sqrt{n}$ is always nonnegative, and u'll have something like $\sqrt{n} < stuff$ after the rearranging we just did, then you also know "stuff" is positive, so that justifies the squaring step
992qqoloy
But for the second inequality you'll have something like $stuff2 < \sqrt{n}$
992qqoloy
So you gotta justify why "stuff2" is positive
Oh wait hmm I just realized I misread
Is it $a < b$ or $b< a$ ? Cus the former means squaring both sides won't necessarily work as you'll have $a - b + 2c$ be "stuff2", so since then $a - b$ is negative, depending on the value of $c$ it could be positive or negative
I've got it now I think!
992qqoloy
so the issue is to go backwards from the result to a known true statement, we have to square the inequality. But that step isn't always reversible right and thats the issue
Yeah
Oh wait I think I know how to fix this
U have two cases: one where $a - b + 2c$ is negative and one where it's nonnegative
992qqoloy
right
In the negative case, clearly $a - b + 2c < \sqrt{n}$
992qqoloy
So u can assume it's nonnegative when u square
valid. Thank you bro!
np
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hello, can you help me understand the approach to this problem
be more specific. where do you get lost first
Ok so i just understoof the first and second line
in this is where the equation came from
now im trying to understand the mathematical process after the second line
like how did we get this
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To be clear is Vy at the first second 9.81m/s?
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To be clear is Vy at the first second 9.81m/s?
And initial velocity for Vy is 0
yes that is correct
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How to do this problem?
whats the question?
$$ \text{ if } \lvert x - 1 \rvert < \delta \text{ Then } \lvert x^2 - 1 \rvert < \frac{1}{2} \text{ f(x) = x^2 find delta so that epsilon = 1/2} $$
Nerdy_Coder
$$ \text{ if } \lvert x - 1 \rvert < \delta \text{ Then } \lvert x^2 - 1 \rvert < \frac{1}{2} \text{ f(x) = x^2 find delta so that epsilon = 1/2} $$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.57 ...x) = x^2 find delta so that epsilon = 1/2}
$$
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```
If $\lvert x - 1 \rvert < \delta$ then $\lvert x^2 - 1 \rvert < \frac{1}{2} \text{ f(x) = x^2$. Find $\delta$ so that $\epsilon = 1/2$.
lmao
∀ϵ>0,∃δ>0
s.t. if 0<|x−1|<δ
, then |x2x+2−13|<ϵ
.
|3x2−x−23(x+2)|=|(3x+2)(x−1)3x+6|<ϵ
Let δ=13
. Then, |x−1|<13⇒|x|<43,|3x+2|<6
..
Then, |(3x+2)(x−1)3x+6|<|6(x−1)6|<ϵ
|x−1|<ϵ
, so δ
=min{1/3, ϵ
}
i think this feels correct, but to be sure... i thank in advance!
ig
its correct
I can’t figure this out, so here is the work i’ve done
I tried working out what delta is, but no method worked
Need more readable text and tell what is part of the question and what is not instead of everything being written as white
$$ \text{ if } \lvert x - 1 \rvert < \delta \text{ Then } $$ $$ \lvert x^2 - 1 \rvert < \frac{1}{2} \text{ $f(x) = x^2 find delta so that epsilon = 1/2} $$
Nerdy_Coder
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
Still broken latex
Get help in #latex-help
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hi can someone help? i tried drawing a diagram
its 80 tho
according to the answer scheme
might wanna redraw the diagram
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determine the vertex form for the function with x-intercepts of (-5, 0) and (-3, 0) that goes throuhj point (-2, -9)'
Please don't occupy multiple help channels.
what have you tried?
I'm assuming you're looking for a quadratic
yeah. Ive tried to find the equation, i got (x+5) (x+3)
i also found the axis of symmetry which is -4
ahh okay, you found most of the equation
but you're missing the amplitude, a
$y=a(x+5)(x+3)$
tatpoj
do you see where you could go from here? @polar ermine
do i distribute "a" to (x+5) and (x+3)
no, I think that'll just make it harder to find a
See how, besides x and y, a is the only unknown in here
so if you have something you can plug in for x and y, you can solve for a
yep
you want it specifically in vertex form, right?
yep
what does your vertex form look like? I don't know of a parameter called q in vertex form
y=a(x-p)^2+q
-9=-3(-2+4)^2+q
yes, that'll work
yep
so y=-3(x+4)^2+3
ye lol
everything seems to work
thx man
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quick question
I can't divide 0/0 right? so the second part is wrong
and the answer should be y' = 1 or y' = -1 and any T thats not t = 0 or t = -2?
is a solution?
This is a question about existence and uniqueness btw
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Could anyone help with this ?
$|2i \cos(2t) - 2 \sin(t)| = 2 \sqrt{\cos^2(2t) + \sin^2(t)} = 2 \sqrt{1 - 2 \sin^2(t) + \sin^2(t)} = 2 |\cos(t)|$
Ann
How did you find the first equality ?
$|x + iy| = \sqrt{x^2 + y^2}$?
Ann
and i also factored the 2 out for convenience.
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23 + x = 69 what's x
!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
How old're you?
do you know basic algebra?
I don't see how that relates with my question
Yes
Okay, then answer Ann's question
ok so then do you know such basic moves as "subtract ___ from both sides"?
Why would we do that
to get x on its own? cause that's our goal, isn't it?
what should we subtract from both sides so that the 23 that's being added to x disappears?
(23+x)^2 = 69^2?
this is like seeing a small pothole in a road, and saying "What if we blew it up with TNT and then put 200 truckloads of concrete into it?"
are you saying we shoulda took cubes on bs
fuck no that's nuking it
Regardless there's one more question
why are you trying to make this equation more complicated than it needs to be lmao
!original
Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.
If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.
if you want you can raise it to the power of n
x^x^3 = 36
Perfect
$x^{x^3} = 36$?
Ann
also uh
Yes
Are you essentially trying to troll @lilac flame ?
bruh
I am not Muhammad
Nah this dude trolling
ok then can you please take a screenshot of the question
How would you go about solving x^x^3 = 36
you would not
I don't have the book okay
Why
I grab desmos
I have only today
this equation is impossible to solve with simple algebra
or some enumerator
Use complex then
also how do we know you did not misremember the question
or microsoft math
Can i show my procedure?
or something starts with G, with a P in middle and T at the end
fuck fof
off*
not that
NEVER USE CHATGPT FOR MATH.
Ikr cant even solve basic binomials
what if it worked
hm
fakeryc please move elsewhere with your AI advertisement
no
happy coincidence, i guess?
Fuck you ik its correct
fun little status u got there too
you could not solve x^x^3 = 37 this way.
i didn't say it wasn't
True but not a question innit
I'd just grab python
Well lovely talk wasn't it fellows
bro have some respect she is helping
make a enumerator
Mb
not really no
Ann senpai im sorry
yeah it was not lovely at all
Thankyou for the trolling you did.
this is the question i was asked but didn't have the balls to say this
if you already had the solution, that you knew was right you didn't really need to come here for that
Why tf is the entire server here
@lilac flame are you over or under 13 btw
Wanted to know about my procedure
Does it matter? I am 16
We are gathered here today to let you know mista V that You Suck
we get underage people in here sometimes
and the difference in level of math compared to your original question
makes people queestion whether your intentions are genuine
^
why open with the most elementary equation possible
Tbh everyone is here.
u need to take a chill pill mate
why the bait and switch
there's no point in it other than to waste people's time and troll
My sir used to say harder questions makes you harder
of whether you're genuinely here to get proper help
or just troll/waste peoples time
ok
pause
bro what
sure
I believe i found the my answer
Do you have any other questions that are genuine and not troll messages?
go whenever, you don't need our permission to leave
Bro what?
In fact yes
Actually yes you can go now
I don't want questions, I meant if you needed any help with more questions
{a} what kind of matrix is this is it a scalar, diagonal, square, rectangular
Where is the matrix?
in his mind
Um its the first word
mistav you are trolling
fs
[a]
the guy is a troll
Alright now better
high level troll
NO I AM NOT ISTG
1x1 matrix is all of the above lololo
If u had to pick one option
ohh wait
Its a scalar matrix first of all.
Nope
Brb
its only nonzero entriesareon the diagonal
it's diag as well?
its also square because 1=1
It's also a square matrix
its also rectangular because every matrix is rectangular
it's a quadruple matrix 
quadrilateral matrix
The correct answer was diagonal btw
But anyway this is a help channel

Am i in general chat?
Hey dont ask me how it is
you're in your own help channel
Idk its an entry question our sir asked
exaplaib
probably
Cuz every diagonal matrix has to be square
why
you are in general troll my friend
I hate people who are all about their sir.

Whats a scalar matrix, Mistav?
mostav can u tell what a scalar matrix is
It can be that but it wasn't
wth is happening here
Who is stopping u just close the channel
Wym it wasn't
STOP ASKKNG MY questions 😭
multiple of identity
Disrespectful innit yall my seniors
Actually we need a modping tbh
do .close
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<@&268886789983436800> read the help channel please.
aight innit
its so intuitive if thats notan actual term then it ought to be
Let them decide
why ping mods lmao
are you our hostage or something
yo mods we probably got a under 13 year old here
Hes not under 13
man i'm getting 11 year old vibes from him
hes just having his fun at the expense of others
wrong reply sorry
hes kindachillin
i was like this when i was 12 👀
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I was wondering what that was
close
lmao
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i am not getting vibes at all
stop talking
Ive never ever created a help channel

were you born in the form of a helper

This is a bug
My message is before the available message
anyway vacate
my internet is trash images are taking forever to load
Please don't make modpings a guessing contest. State what it is you think needs moderator attention.
false
You tell em mod senpai
Btw i am 16
Well. Uh I meant read the start of the help channel... The questions seem all trolling.
You just got told off brother
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Closed by @agile trellis
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How can the question be simplified to only sqrt of 5
can someone explain what this means
yh sos
np
Can someone explain what the question is asking for?
surely all integrals go to something
uhhhhhh
lemme search that up
lol i dont even know what all 3 is
how does that answer my question? slightly confused
it dont
i cant sorry, but someone will eventually help you
@nimble ibex Has your question been resolved?
<@&286206848099549185>
can you explain what this question wants and how to do it?
thanks
It's saying, is the area finite
so yes because sin(0->90) is never 0
Yes
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Could someone help me with 2 b ? This is what I got for 2 a
Haven’t you got help earlier?
Oh i see
do you see, x being a variable can result in different area based on the input for x from part a?
4x+4 is rather an equation for finding area when you have a varying length x
do you see that ?
Ye
So for different values of x you get different area right?
or for a specific area you have the corresponding length specific value for x.
now the area of the figure being 18
what does that tell you?
you have to set the length accordingly so the the total sum becomes 18 right?
or 4x+4= 18
Ye
better to subtract 4
first
your goal is make x alone in the right, by undoing the process
your x is being multiplied by 4 and added 4 leaving you 18
now you undo adding 4 by subtracting 4
4x = 14
yes, now undo multiplication
X = 7/2 ?
sure
Is that the final answer of the question ?
What is the question asking you to find?
X
and what have you found?
X !!
Good job
Cheers 👍
hf
Do you mind helping with another question ?
Claim a new channel and post
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Find all angles and sides of a right triangle, knowing angle B and the sum of the cathetus (AB + AC)
so far I have used pythagora's theorem, this way:
beautiful man
i am stuck from this point
i thought maybe i can write b^2+c^2 as (b+c)^2 - 2bc
but sadly the product is not given
beautiful man
<@&286206848099549185>
@crude lark Has your question been resolved?
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what are you given?
i solved it
oh nvm
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how do i do this?
!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
2
!show
Show your work, and if possible, explain where you are stuck.
wait no 3 i mean sorry
-3x+18≥6
18≥6+3x
12≥3x
4≥x
Its correct
ah
i must have typed -4 on accident as thats what a help source had used an example of, sorry for the inconvenience
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Can someone help me get from the thing I underlined in blue to the one I drew the arrow to
I think I'd have to group them somehow but I'm lost
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Hello, excuse me
.close
type this @karmic vortex
.close
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V = span{(1, 1, 2), (1, 0, −1), (2, 3, 1)} What is the base and the dimention?
I can't understand something. For those 3 vectors to be the base, they all have to be linear independent.
How can i check that in a trivial way?
Shouldnt it be like this?
(Uploading pic)
Add equation 1 and 3
can add equations #1 and #3 and get 3α + 3γ = 0
it becomes alpha = gamma, right?
no
yes
Doesn't that make them dependent?
Let me draft something in a clean paper to show you
put α = -γ into α + 3γ = 0 and you will get γ = 0 and from this the other two letters will also turn out as 0
Yeah I was getting into that
Please give me a minute to show my result
I jumped a few steps since, I suppose that, knowing that alpha and gamma are 0, using the first equation with that information, beta will be 0 aswell
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@random veldt Has your question been resolved?
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I need to ask about using the fourier transform to solve the heat equation
I'm working with $$\phi _t = \kappa \phi _{xx}$$
jan Niku
I'm able to get this with forward transforms and integrating factor to $$\pdv \qty( \hat \phi e^{\kappa n^2 t} )$$ providing $$\hat \phi (t,n) = \hat \phi _0 (k) e^{-\kappa n^2 t}$$
jan Niku
now I'm confused about the reverse. The thing I'm following says that we want some $S(x,t)$ such that $\mathscr F ^{-1} { \hat S(n,t) } = S$
jan Niku
this by the definition of the fourier transform im using should mean $$S(x,t) = \frac{1}{2\pi} \int _{\bR} e^{-\kappa n^2 t}e^{-inx} \dd n$$
jan Niku
I'm having a hard time with this integral is the hard part. Mainly, getting the same thing the guide says
we would just be able to complete the square, right?
so $\frac{1}{2 \pi} \int _{\bR} e^{ \qty(\sqrt{\kappa t} n + \frac{ix}{2\sqrt{\kappa t}} )^2 + \frac{x^2}{4\kappa t}} \dd n$
jan Niku
using $\int _{\bR} e^{a(x+b)^2} \dd x = \sqrt{ \frac{\pi}{a} }$, shouldnt this be $$\frac{1}{2\pi} e^{ \frac{x^2}{4\kappa t} } \sqrt{ \frac{\pi}{\sqrt{\kappa t}}$$?
jan Niku
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send your work
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Closed due to the original message being deleted
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Ok
Sorry
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You're going to have to wait for this channel to fully close
ight
nah you can open a new one iirc
They can't open a new one if there were none available
there's no channels avaliable, damn funny asf ngl

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Hello, would someone be able to explain this to me:
You buy a stove for $1500 on February 20. The store gives you 4 months “interest free”, after which you must pay the $1500. However, the store charges an administration fee of $50, to be paid today. What rate of simple interest, r, are you being charged for this “interest free” plan?
I know the answer but I dont know how it arrived at that. The answer is 10.49%
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how are these integrals solved differently?
i never knew the placement of dx mattered
wdym solved differently
in the second one, only 6x^5 is being integrated
the dx among other things is indicating "end of the integrand"
so everything before the dx is what we integrate?
yea generally
unless you're a physicist weirdo who writes the dx first haha
don't be that
so the first integral would be x^6 - 6x^3 + 7x + C
and the second integral would be
x^6 - 18x^2 + C ?
i lumped the constant 7 into C
or maybe we can just leave it as x^6 - 18x^2 + 7
Me \s
yes
still should have a +C either way and yea, it's nice to lump the 7 in with the C
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You're welcome
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I can not use MCT or LDC
what's up with the limit in the first line, there's no n
yeah its not correct let me reformat it 1 sec (FIXED QUESTION HERE*)
fixed now
ah that's better
are there any hypotheses for X, like is it a finite measure space for example
no
Only thing I have shown is that u is also positive measurable and I have shown that \int_X u d\mu \leq 2023
and that \int_X u d\mu = 2023 if the sequence is increasing using Beppo levi
but (X,\mathcal{A},\mu) is just a mesaure space thats all the info
I was thinking the approach might be to sandwhich the integral as I already has that it is \leq 2023 but the limit of the integral of u_n is equal to 2023?
hmm, offhand maybe first show that it's true if X has finite measure, then see if you can somehow extend that to the general case. If X were sigma-finite you could use a sequence X_1, X_2, ... of measurable sets with finite measure whose union is X, but in the general case that won't work
you could try asking in #real-complex-analysis also
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bbbb
why is roc at minor axis b/a^2 ... if a circle is lying below the curce then isnt roc -ve? and even by formula its coming out to be -ve
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<@&286206848099549185>
.
roc is radius of curvature
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Let's say I have
Ax=B, is it true that A^-1*Ax = A^-1*B => x = A^-1*B
In general, is A^-1*A = I?
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how do i solve this question?
image on its way?
!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
yes you do
but i'm not sure which one
well you might as well start by writing out the equation for its position wrt time
like
$x(t) = x_0 + v_0t - \frac{1}{2}at^2$
Ann
and also $v(t) = v_0 - at$ while we're at it?
Ann
like ok there is probably a pre-derived one
but i don't remember it off the top of my head
v_0 is initial velocity, and x_0 is initial distance, right?
v(f)²=v(i)²+2ad
24²=2ad
a=576/38.4
a=-15ms⁻²
thanks
what about this one?
i think i might need simultaneous equations
but i'm not sure how to implement it
@tropic oxide
you don't really need simultaneous equations here
you can find the rate at which A catches up to B
you mean something like 340/280?
no
if you run at 5 km/h and i run at 8 km/h at what rate would i be catching up to you
3.5 hours
at the speed of aircraft b, 3.5 hours covers 980 NM
ok
thanks
i have 1 more
@tropic oxide
too tired for this shit sry
ideal gas law yes?
then what's the problem
i haven't studied it
not that i was too lazy to study or anything, i was just never taught it to begin with
but you're supposed to know it for whatever this is for?
chlamydia
i'd imagine p is pressure and v is volume
yes
r seems like a constant
T in kelvin
for mols
mols are a measure of the amount of a substance, right?
sorry, i don't do chemistry
yeah constant n and V
wait
i don't need to know n, do i?
so they're not needed
yeah
so oneside changes exactly as much as the other side
i know that 22c in kelvin would be 295.15
p increases by a factor of 2/3
so T would also increase by the same amount, right?
yeah
so it's about 494k?
huh
492
close enough
especially since i can't even use a calculator for this exam
on the bright side, it's only MCQs
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Can you check if the proof is correct
as well as help me with proper way to write it
Ann
other than that it seems fine.
Is the scrarch work written in proof side?
I feel like we do scratch work backwards in proof
I would make it more explicit where you are using the triangle inequality
you should get used to writing a chain of inequalities, instead of inequality implies inequality implies inequality
and the proof should start with "let epsilon>0 be given"
can you elaborate a bit?
|an+bn-a-b|<=|an-a|+|bn-b|<eps/2+eps/2=eps
I see
This arbitrary choice of epsilon is for the sequence (an+bn) right?
yes
I’m making use of things I’m guaranteed from other two convergence
in general, any symbol you use in the proof also has to be introduced somewhere
right
my epsilon came out of the blue
Thanks for pointing it out
Essentially the choice we take from the definition of convergence of (bn) is after we fix epsilion for (an+bn) right?
i think that’s what you mean
yes
for an+bn you have to show it for any epsilon
for bn you can use it for any epsilon
which you do for epsilon/2
u2
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i did part (a) and (b) but need help on (c). i feel like its true but im not sure how to explain it
still try explaining
like for any bounded function where n exists, it is strictly bounded under k = n + 1
similarly, for any strictly bounded function where k exists, it is also bounded under n = k
i dont feel like this is "valid"
it is
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Given there’s 9 balls in a bag, in which 2 of these 9 balls are named number 2, 3 of them are number 3, and 4 of them are number 4. Now we pick 3 balls from the bag, what is the expected value of the number sum?
This question is quite time-consuming.
If you list out all of the possibilities for the numbers of these 3 chosen balls.
However,
My textbook mentiond that the EV would equal to the EV with just one pick from the bag multiplied by 3
And that sounds like a cap to me
I wonder how the clever trick works
once a ball is taken, we do not put it back
The book is wrong
but it is true
Oh it is?
it is
same when i first seen that trick
!original please
Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.
If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.
Still thinking about it, but i just want to be sure
it is not being written in English, but i will send it tho
the book is right

