Elden ring and jedi has same server tag
#help-36
tall breach
tall breach
W spike player
Wt
tall breach
Ops
tall breach
W spike player
I wanted to ask you Wt was the meaning
mellow vortex
I wanted to ask you Wt was the meaning
Wt= what the
tall breach
Then explain why WTS means what the sigma
Well k idk
mellow vortex
Well k idk
idk= i dont know and k means ok
final saddleBOT
@chilly moon Has your question been resolved?
lost zealot
Hello, I am stuck on this question about probability (i think) and some help would be appreciated!
final saddleBOT
Hello, I am stuck on this question about probability (i think) and some help wou...
tired walrus
try to write out all possible roll sequences that would result in a running total of 3 at some point. you may use some placeholder or wildcard like x to represent a roll that could be any number
for example the roll sequence 12xx would fit this description
tired walrus
running total = the sub-totals you get during the rolling process
like if your rolls are 6, 4, 2, 3 you will have the running totals 6, 10, 12, 15
lost zealot
oh
49/216?
the probability of landing it on the first one is 1/6, then 1/18 (second) then 216 (third one)
you add them which equals 49/216
left trail
For this I know that n! = $\prod_{k=1}^{n}a_k$
final saddleBOT
For this I know that n! = $\prod_{k=1}^{n}a_k$
soft zealotBOT
BigBen
left trail
But when I break up the nCk wouldn't I be introducing a new term if I just plugged that in?
tired walrus
no, $n! = \prod_{k=1}^n k$
soft zealotBOT
Ann
tired walrus
not some un-introduced sequence a_k
left trail
not some un-introduced sequence a_k
Sorry misread the book but either way if I just to rewrite that in the binomial theorem you can just sub that in then?
tired walrus
ehhh
well yes i guess thatll happen at one point
feels like a "less talking more doing" moment rn though
left trail
feels like a "less talking more doing" moment rn though
Ye so I ended up getting stuck at either $\frac{1}{(n-k)!}\sum_{k=0}^{n}\frac{1}{k!}\prod_{r=0}^{k-1}(r+1)(\frac{1}{n})^{k}$ or $\frac{1}{(n-k)!}\sum_{k=0}^{n}\frac{1}{k!}\prod_{r=0}^{k-1}(r+1)(\frac{1}{n})^{n-k}$
soft zealotBOT
BigBen
left trail
For this I know that n! = $\prod_{k=1}^{n}a_k$
I am trying to make it as similar as this
There isn't any 1/(n-k)! there
tired walrus
what is 1/(n-k)! doing outside the sigma at all though
i think you may have fumbled this quite severely, but i can't tell where -- aside from the issue i just mentioned.
this is what i wrote down just now though
tired walrus
There isn't any 1/(n-k)! there
im saying YOUR thing has a glaring notation issue
final saddleBOT
@left trail Has your question been resolved?
old quarry
final saddleBOT
old quarry
yeah
eager shore
what happens if you rationalise the denominator
old quarry
we need to get it in form of a-b/1+ab
old quarry
what happens if you rationalise the denominator
we get 2r
but the numerator becomes some ugly shit
eager shore
Oh look
soft zealotBOT
dyxn
eager shore
(1 - ab)/(a + b)
eager shore
multiply the top and bottom with (1 - r^(-1))^(1/2)
Then you can split it
I think
💀
old quarry
ok let me try
old quarry
multiply the top and bottom with (1 - r^(-1))^(1/2)
why will we multiply by this btw?
eager shore
it gets it in the form (a + b)/(1 - ab)
kind of weirdly though
but it does
i think you can fuck around and make it telescope
old quarry
multiply the top and bottom with (1 - r^(-1))^(1/2)
oh ok yeah this works
got it
thanks
.close
final saddleBOT
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ionic gate
Is sup not about the interval but about what is the least upper bound of the given interval?
final saddleBOT
Is sup not about the interval but about what is the least upper bound of the giv...
ionic gate
So, this would have an upperbound of u...infinity?
Is this not the upper bound because it has no least upper bound?
versed crater
I don’t get what you’re asking
versed crater
Is this not the upper bound because it has no least upper bound?
There’s no the upper bound because it’s not necessarily unique
final saddleBOT
@ionic gate Has your question been resolved?
tranquil pine
if we have a problem like this (20) can we assume that different variables mean different values?
final saddleBOT
if we have a problem like this (20) can we assume that different variables mean ...
coral prawn
if we have a problem like this (20) can we assume that different variables mean ...
Different variables can be equal unless otherwise stated
final saddleBOT
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Available help channel!
kindred sun
final saddleBOT
kindred sun
i got 4 seconds
robust mulch
!redir
final saddleBOT
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
kindred sun
This channel is only for on-topic discussion. Please take casual conversation to...
sorry
barren hound
i dont have a girlfriend
do you want one?
kindred sun
sure
tired bramble
hello, i am having some difficulty understanding this problem in genral
tired bramble
and kinda forming a counter example. after some thinking i have come to the conclusion that 1. false as if a function is bijective then not a subset of it should be bijective but all of it should be bijective
so given this 1. is false and 2. is true
this is my ounter example for 1. i think it works
but i have no idea how to start on B
$\forall A \subseteq \mathbb{R},\quad f^{-1}(f(A)) = f(f^{-1}(A))
\quad \Leftrightarrow \quad f \text{ is bijective}.$
soft zealotBOT
SushiMan
tired bramble
for counterexamples the empty set is handy because it's so simple
how does that work
also does my counter example work or nah
wait
nvm
ripe jewel
f o finv{varnothing} = varnothing = finv o f (varnothing) vaciously
it doesnt tell you anything at all about f
tired bramble
tired bramble
f o finv{varnothing} = varnothing = finv o f (varnothing) vaciously
😭 ts so hard to read without latex gimme one sec
ripe jewel
but anyway it's asking about f: R to R I just realized
ripe jewel
$f \circ f^{-1}(\varnothing) = \varnothing = f^{-1} \circ f (\varnothing)$
soft zealotBOT
gfauxpas
ripe jewel
but its a subset
but a different subset for domain and codomain so not what the problem is asking about
also read the question carefully
it's asking for sufficient and necessary conditions, not equivalences
tired bramble
**gfauxpas**
thats vacously true tho, so it doesnt work as a counter example
ripe jewel
thats vacously true tho, so it doesnt work as a counter example
it shows that a is not sufficient for anything
because it's true for every function
tired bramble
im really ocnfused ngl
ripe jewel
okay so we're dealing with 4 questions here
tired bramble
ok
ripe jewel
is (a) necessary for bijectivity? sufficient?
is (b) necessary? sufficient?
I'm saying that my example with the empty set shows (a) cannot be sufficient for anything, because it's true for every function ever
that knocks out 1 of the 4 questions
tired bramble
What does sufficient mean again
ripe jewel
$p \implies q$ is "p is sufficient for q"
soft zealotBOT
gfauxpas
ripe jewel
and "q is neccesary for p"
think about why it's called that in terms of regular english words,it's not arbitary
tired bramble
Oh so if it’s nesscary then it’s the other way around
ripe jewel
the question is asking about the image of sets under f. Can you convert that into a statement about images of elements under f?
LMK if you dont know that terminology, image under
tired bramble
Reverse direction
Bro I got this question wrong on the test and it cost me a whole percent 😭
Ok alr lemme think
ripe jewel
Do you know that terminology, image under?
tired bramble
No I don’t know
ripe jewel
if A is a set,
tired bramble
But it is nesscary but it isn’t sufficient for a
ripe jewel
f(A) is called the image of A under f
if x is an element in the domain of f
then f(x) is the image of x under f
ripe jewel
so if you're told the equality is true for EVERY subset
is there a collection of subsets that tells you information about how f and f inverse act on ELEMENTS?
tired bramble
That’s for b no
ripe jewel
they're related
tired bramble
is there a collection of subsets that tells you information about how f and f in...
Power set
That’s for b
ripe jewel
singletons
tired bramble
I’ve never heard of that
ripe jewel
sets with exactly 1 element
tired bramble
Ohhh okay
ripe jewel
every number r in R corresponds to a singleton {r} in the powerset
so if you're told a statement is true about every subset
in particular it's true for every singleton
tired bramble
He’s
ripe jewel
He's indeed
tired bramble
Sorry I was tryna say yes
tired bramble
so if you're told a statement is true about every subset
Every subset yes
That’s basically B
ripe jewel
so b is sufficient, or necessary, for invertibility?
ripe jewel
yeah it's both but i only proved it's sufficient actually
but that also means a is necessary
tired bramble
Wait for b right
ripe jewel
yeah for b
tired bramble
but that also means a is necessary
Yes for a is nesscary
I stated that earlier. W prediction
ripe jewel
the way youd prove b is necessary is because
inverse image of a union is the union of the inverse image
therefore... .can you finish the proof?
tired bramble
therefore... .can you finish the proof?
Probably not
Tbh I kinda get what ur getting at but it’s still pretty confusing
ripe jewel
if it's true for all singletons, it's true for all unions of singletons
which is, all subsets
ripe jewel
Tbh I kinda get what ur getting at but it’s still pretty confusing
sure
I've been doing this for years
takes practice, and looking at pictures sometimes
tired bramble
I've been doing this for years
Are you a prof?
tired bramble
every set is a union of all singletons of stuff in the set
W thinking
ripe jewel
no, I am an out of work math tutor studying to pass actuary exams
tired bramble
What are those
tired bramble
every set is a union of all singletons of stuff in the set
Yes this makes sense
So I understand this. What are we getting at now
ripe jewel
they are standardized tests to show that you are proficient in the types of math insurance companies need to do
tired bramble
they are standardized tests to show that you are proficient in the types of math...
Oh I see
What kinda job are you trying to get
ripe jewel
work for someone who is working for insurance companies (an actuary), most likely. easier than becoming an actuary myself
but honestly ill take any math job atm , im so poor :(
tired bramble
work for someone who is working for insurance companies (an actuary), most likel...
Oh I see
tired bramble
but honestly ill take any math job atm , im so poor :(
You can’t do ta ship?
I’m first yr but I have like a lot of third year friends that are tas
They get paid 53 an hr
53 cad btw
ripe jewel
usually that job is given to PhD students, no?
tired bramble
Nah
Utm cs has 3rd students as tas for first yr courses
Uft cs in general I think
ripe jewel
yeah but students still. i graduated already
tired bramble
yeah but students still. i graduated already
Iirc you can still be a ta even after you graduate. One of my head tas had a faang job and was still doing ta ship for a first yr course
tired bramble
What kinda job are you trying to get
Anyways back to the topic. What are you trying to get at here tho
@ripe jewel
Are you French by any chance
J’ai faux pas
ripe jewel
no, faux pas has entered English
as a borrowed word
i forgot what we're talking about
final saddleBOT
@tired bramble Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
hearty mica
We are given the point of a final side of angle Theta. Find the values of all trigonometric functions of theta.
final saddleBOT
We are given the point of a final side of angle Theta. Find the values of all tr...
hearty mica
Im trying to do (2, - 3)
My first thought is to find theta like this:
cos(theta) = 2
theta = arccos(2)
But arccos of 2 isnt a thing
So I dunno how to find that angle
blissful meadow
How did you get cos(theta) = 2?
cos(theta) is adjacent / hypotenuse.
Have you found the hypotenuse?
Moreover, although it's not wrong per se to solve for theta, you realistically don't need to know theta in order to compute all the trig ratios associated to it...
hearty mica
How did you get cos(theta) = 2?
X is equal to the cosine of the angle, and in this case 2 is x
hearty mica
Moreover, although it's not wrong per se to solve for theta, you realistically d...
Explain
blissful meadow
X is equal to the cosine of the angle, and in this case 2 is x
Perhaps, but cos(theta) is not just x
It's x over the hypotenuse
blissful meadow
Explain
Perhaps the easiest way to ease into it would be to draw a diagram
Using pythagoras you can find the hypotenuse
hearty mica
Using pythagoras you can find the hypotenuse
I see
I think im pikcing up what youre laying down
I get the hypotenuse, and then the functions are just the ratios of all sides,
blissful meadow
This approach, which is likely the one they want you to use according to the wording of the question, avoids having to deal with issues with the domain and range of the inverse trig functinos
hearty mica
Yeah we havent gotten into that yet
I'm still confused about what i was told though, about x being the cosine of the angle
We're working with circles if that helps
blissful meadow
It is the cosine of the angle in the very special case where the hypotenuse is 1, i.e. on the unit circle!
hearty mica
I see
But in this case, the coordinates are outside the bounds of the unit circle
Right?
I see, in the unit circle, the hypotenuse is 1, ie, cos(theta) = x / 1, ie, cos(theta) = x
.close
final saddleBOT
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prime tusk
sullen gale
Hey!
sullen gale
Do you know how to calculate the area using the inradius @prime tusk ?
storm haven
prime tusk
Do you know how to calculate the area using the inradius <@1104619287206952960> ...
i dont hink os
sullen gale
look at the @bove image
sullen gale
@prime tusk this one
prime tusk
is there any other way to do it or something
idk i feel like my teacher will think i cheated or something
if i do that
prime tusk
<@1104619287206952960> this one
^^
sullen gale
@prime tusk Has your teacher taught you this and/or has it appeared in your textbook?
sorry for the late reply
you can also derive it like they did in the image
like split it into three triangles and work it out
however
I think if your teacher gave you this problem, it must've already been taught
final saddleBOT
@prime tusk Has your question been resolved?
prime tusk
I think if your teacher gave you this problem, it must've already been taught
idk he pmo
he teaches me like
sullen gale
alright then. So you can do a similar thing that's shown in the diagram, splitting it up into 3 triangles and using variables to calculate the area
prime tusk
alright then. So you can do a similar thing that's shown in the diagram, splitti...
doesnt that get me nowhere
bcz i dont know a b or c
sullen gale
that's fine, just keep them as expressions in terms of AB, AC, and BC.
can you calculate the top-left triangle?
the area
prime tusk
uhh
prime tusk
can you calculate the top-left triangle?
um which one
sullen gale
using your original diagram
prime tusk
the one on the top or the one on the left
sullen gale
in terms of r and AB
sullen gale
split it up like this one
prime tusk
split it up like this one
sorry uhh
im still confused
bcz
i have
idk
nothing almost
sullen gale
can you calculate the areas of all three triangles in terms of their side lengths and the radius?
@prime tusk
prime tusk
but i dont know the sidelengths
sullen gale
that's alright, do it in terms of them
let me do the first one as an example
the area is base.height/2=AB.r/2
now can you do the same for the other triangles?
prime tusk
but thats js
sullen gale
no trust the process
prime tusk
30
prime tusk
r(AB+BC+AC) divided by 2?
sullen gale
yep. do you know these values?
prime tusk
.. no
sullen gale
you know r and AB+BC+AC right?
prime tusk
no...
sullen gale
isn't it given in the problem? look at your pic
sullen gale
yeah 😄
prime tusk
DUDE YOUR A GENIUS
sullen gale
😉 good luck on your hw!
dense temple
Did someone know how to prove this geometrically? The text says ABCD is a square and that we should prove that expression.
final saddleBOT
Did someone know how to prove this geometrically? The text says ABCD is a square...
dense temple
By the way, it occurred to me to use sine and cosine of 127/2 and find a and b squared, but I have no idea how to relate it to L squared, I don't want to complicate things so much
final saddleBOT
@dense temple Has your question been resolved?
final saddleBOT
@dense temple Has your question been resolved?
final saddleBOT
@dense temple Has your question been resolved?
halcyon swan
can someone help me understand this particular problem
final saddleBOT
can someone help me understand this particular problem
halcyon swan
magic coyote
What part
halcyon swan
like after the substituting the x and y respectively then having this "2D+3E+F =-13" and so on
magic coyote
Mhm
magic coyote
Ok after that it's just another system with 2 variables and 2 equations
4D + 2E = -20
-D-E=12
Can you solve this for D and E?
halcyon swan
uhh
halcyon swan
Can you solve this for D and E?
what are we solving?
magic coyote
Those 2 equations, for the variables D and E.
dapper hull
E= -10 -2D
D=-12-E
you still haven't found either D or E so this won't help
halcyon swan
ah ill try again
dapper hull
you still haven't found either D or E so this won't help
correction: substitution using either would though
dapper hull
yea it's correct
halcyon swan
yay
dapper hull
lowkey that's my first time encountring something like finding the conic equation using 3 points lol
dapper hull
you need to use all cuz in most case 3 equations are needed to solve 3 variables (in your case D E F)
and all 3 points should work (lie on the curve of the conic
dapper hull
at least F would always be F lol easy to get rid of
halcyon swan
ye
dapper hull
you got F yea ?
halcyon swan
working on in now
dapper hull
just sub in E D (use any of the 3 equations)
halcyon swan
i got 25 on the first one
halcyon swan
just sub in E D (use any of the 3 equations)
yesss thanks
dapper hull
indeed it should work (I checked the answers using desmos)
All 3 points lie on the conic (circle)
harsh solar
Are the eqns i formed correct for qn13?
final saddleBOT
Are the eqns i formed correct for qn13?
@harsh solar Has your question been resolved?
final saddleBOT
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Available help channel!
hushed dust
Hello, can someone help me with this problem, please?
It is known that, under optimal conditions, a certain machine produces screws with a diameter of 0.5 cm and a standard deviation of 0.02 cm. This machine requires periodic adjustments, but those are costly. Therefore, a daily sample of 100 randomly selected screws is tested, and the machine is adjusted only if the average diameter of the screws exceeds a certain value x cm. Find the value of x such that the probability of adjusting the machine when it is not necessary is at most 5%.
final saddleBOT
Hello, can someone help me with this problem, please?
It is known that, under op...
sullen gale
!show
final saddleBOT
Show your work, and if possible, explain where you are stuck.
hushed dust
I'm having trouble interpreting the statement. Is the value 0.5 supposed to be the mean (μ)?
sullen gale
Doesn't it say that it's 0.5 cm? And that's the diameter of the screw.
hushed dust
But it does not explicitly state that it is the mean.
hushed dust
hmm..
sullen gale
It tells us that the standard deviation of the screw diameters is 0.02 cm
-# do you have anything to add, @dense badge ?
hushed dust
Is this the hypothesis test?
H0: μ<=0.5
H1: μ>0.5
dense badge
but you won’t test the hypothesis they want the rejection point x
two sided or one sided?
hushed dust
one sided
hushed dust
but you won’t test the hypothesis they want the rejection point x
this is correct?
dense badge
this is correct?
assuming normality of the observations i think yes, you used the sampling distribution of the mean when obs are normal i think
hushed dust
assuming normality of the observations i think yes, you used the sampling distri...
But since n = 100 > 30, by the central limit theorem, the sample mean tends toward a normal distribution.
dense badge
yeah i think you are good
hushed dust
thanks
tranquil pine
final saddleBOT
<:really:995844031017910342>
tired walrus
<:really:995844031017910342>
hello, do you have a math question to ask?
raw merlin
!xy
final saddleBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
raw merlin
I cannot understand the question.
tranquil pine
stuck juniper
Ur on ur own with this one 🥶🗣️
lyric obsidian
Not saying I can answer the question, but what exactly are you having trouble with understanding this proof?
Also, what is Lemma 4.3.6, Theorem 4.2.8, etc.?
tranquil pine
sullen gale
!occupied
final saddleBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
tranquil pine
-# See bot's message
Aghhh 🤦♀️
final saddleBOT
@tranquil pine Has your question been resolved?
tranquil pine
warped delta
Hi, I'm new here and I want to get back into maths and I'am awfully confused on where to begin.
final saddleBOT
Hi, I'm new here and I want to get back into maths and I'am awfully confused on ...
stone wagon
being more specific would help
warped delta
Like where would you recommend someone who never got through high school fully on where to begin again on learning maths essentially
stone wagon
khan academy has a lot of resources
tardy void
What do you know
warped delta
honestly have no clue about my level
stone wagon
do you know what like idk sin(x) is
warped delta
a little
warped delta
What do you know
not much
I know the basics additon, subtraction, multiplaction, etc and some fractions
Kinda forgot abt the rest
stone wagon
and what is your goal in learning math?
do you wish to get to like a high school level?
stone wagon
then khan academy is perfect for this
warped delta
Alright thanks for the help
sullen gale
You should check out Art of Problem Solving as well
warped delta
is that a book?
stone wagon
You should check out Art of Problem Solving as well
i am not sure if that's for him rn
sullen gale
videos tho?
stone wagon
hm haven't seen their videos
sullen gale
not for like extremely hard contest problems
stone wagon
alr
stone wagon
is that a book?
website
warped delta
I should check them out?
stone wagon
also
https://openstax.org/subjects/math
bunch of free textbooks if you like that approach more
stone wagon
luckily for you, for pre-university math there's a TON of resources online out there and i wouldn't necessarily call anything bad either
so just google stuff or ask questions here if you need more assistance
stone wagon
Thank you both 🙏
if you don't have any more questions, type ".close"
sullen gale
however I do think Khan academy might be best, AoPS has limited videos for material above algebra
warped delta
kk
sullen gale
best of luck on your math journey!
final saddleBOT
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Available help channel!
tranquil pine
i was wondering what was going on in this problem
final saddleBOT
i was wondering what was going on in this problem
tranquil pine
i thought the probability would be 89^2/100^2
since all the divisors of 10^88 are also divisors of 10^99
loud sundial
I don’t see how that assertion gives you that answer
tranquil pine
I don’t see how that assertion gives you that answer
so since probability is favorable/total
OH WAIT
oops i get it now nevermind
.close
final saddleBOT
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Available help channel!
alpine prawn
Question 10 I'm seeing true on a graph i sketched (anwer is false)
Do u have anything to comment on it
final saddleBOT
Question 10 I'm seeing true on a graph i sketched (anwer is false)
Do u have any...
alpine prawn
vital crag
If it's false, find an element in one set that's not in the ither
alpine prawn
.close
crisp mason
final saddleBOT
terse crypt
what u have tried
final saddleBOT
@crisp mason Has your question been resolved?
dense temple
Can anyone find a logical sequence for this?
final saddleBOT
Can anyone find a logical sequence for this?
storm haven
What have u done?
tired walrus
well if you write it as
3/7, 6/12, 9/17, 12/22, ...
then something might come up.
dense temple
What have u done?
I set the number of terms as N and played with it in the dominator and numerator to find a general form
storm haven
Well just look at what Ann said
dense temple
well if you write it as
3/7, **6/12**, 9/17, **12/22**, ...
then _something_ m...
plus 3 in the numerator and plus 5 in the denominator
thanks!
.close
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tulip depot
Hi guys
tulip depot
can anyone help with physics
pls
calculate the final velocity of a 60.0kg mass moving at 2.67 ms*' east, when a net force of 45.5N west acts on it for 2.80 seconds.
tall portal
dont you think you should use that time and force for a quantity
tulip depot
I put the wrong quesiton before
tall portal
it still is the same with diff values
tulip depot
calculate the final velocity of a 60.0kg mass moving at 2.67 ms*' east, when a n...
for this i get to do f=ma
but i got a positive answer but its supposed to be a neg accelleraton
but idk why
tall portal
you took
the force as -ve?
dont you think the directions would point along the same axes
leaden moon
it's hard finding problems that i can solve in help in this server 😭
tall portal
it's hard finding problems that i can solve in help in this server 😭
its alright
tulip depot
wait so this is what i did first :
F=ma
A= f/m
A= 45.5/60 = 0.76ms-2
And then from there im supposed to put that into vf=vi+at but when i did that with the positive acceleration it was wrong, and when i did it with the neg acceleration its right but idk why
tall portal
wait so this is what i did first :
F=ma
A= f/m
A= 45.5/60 = 0.76ms-2
And the...
here
you took
leaden moon
its alright
like one time someone asked for the factorial range
leaden moon
and since west and east are in opposite directions
then if you took east positive
then a would be negative
idk this kind of physics 💀
tall portal
yeah you are right
leaden moon
i learned archimedes formula but not F = ma 💀
tulip depot
arggghh!!! This is making me crash out, the velocity is east but the force is west but why is the acceleration neg
like how do i know its supposed to be neg
im so sorry this isnt making sense to em
leaden moon
well then your object is slowing down
tulip depot
me
tall portal
arggghh!!! This is making me crash out, the velocity is east but the force is we...
you appplied force in west? if you take west positive that is f positive wont a be positive aswell
leaden moon
calculate the final velocity of a 60.0kg mass moving at 2.67 ms*' east, when a n...
F = ma
is 2.67 ms^2 acceleration or velocity?
tall portal
tulip depot
not final
leaden moon
then what's the acceleration 😭
tulip depot
0.76 but idk why its neg and not pos i mean i do but i dont
tall portal
0.76 but idk why its neg and not pos i mean i do but i dont
did you not take f positive while doing f=ma
that meant
leaden moon
F = ma
tall portal
you took west positive
tall portal
hence a will be positive
tulip depot
I’m so confused I’m so sorry
tall portal
its all good
tall portal
if you took f as positive while doing f=ma that implied that you have assumed west direction to be positive
if you get a as positive it means accleartion is along west
tulip depot
if you took f as positive while doing f=ma that implied that you have assumed we...
Yh, bht when I put that in my calc i got 5 Smt and the marking keh said it was wrong
tulip depot
Yh, bht when I put that in my calc i got 5 Smt and the marking keh said it was w...
As in like
tulip depot
As in like
The whole completed answe
tall portal
Yh, bht when I put that in my calc i got 5 Smt and the marking keh said it was w...
it is because in the general ways we assume that east is positive
and west is negative
hence if you take
f in the general coordinates it will be negative here
you get a as a quantity you want
leaden moon
yes
so anyways
you got the acceleration
then i assume you can calculate the final velocity
leaden moon
is there any integral involved?
tall portal
..
tall portal
no you're good
tulip depot
Thank you!!!!
leaden moon
np
tall portal
yeah just .close unless you need anything else
tulip depot
You are the best!!! Saved my life man
leaden moon
how do you calculate the final velocity if you know the acceleration and the initial velocity?
tulip depot
Vf=vi+at
leaden moon
do you use integral for smooth calculation?
tulip depot
Uhhh
tall portal
i am pretty sure accleartion is a constant hence its a hint
leaden moon
alr
final saddleBOT
leaden moon
np
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Available help channel!
devout breach
final saddleBOT
devout breach
i don't have the time to try every single number
lofty willow
Ignore me.
pliant shore
i don't have the time to try every single number
contrary to the deleted message, 4 is indeed possible
pliant shore
can you figure out how? you can try a few possibilities
or even simpler, 0.5 * 8.0
but yeah
I mean 8.0 is technically a decimal cause sig figs matter
not sure what the question intends but in any case
yeah and similarly 4.8 = 0.5 * 9.6
devout breach
so A is wrong
lyric obsidian
I think this is a sig fig thing
magic coyote
Every single product is possible
lyric obsidian
No
pliant shore
so notice how you can take double of 4 and 4.8 and it works
lyric obsidian
I can't tell
devout breach
wait let me make it a.5 x b.c
lyric obsidian
But, what is the maximum amount of digits in a 2 digit by 2 digit multiplication?
pliant shore
it worked with 4 * 2 = 8 and 4.8 * 2 = 9.6
devout breach
4
magic coyote
oversight lmao
pliant shore
clever problem btw
devout breach
99 x 99 = 9801 so it can be 4 digits
lofty willow
You're limited by x.5 and a.b right?
magic coyote
it is
silent cove
as said, you for 4 you can do 0.5 * 8.0
silent cove
for 4.8 you can do 0.5 * 9.6
devout breach
hodl on let me think
pliant shore
yeah then by the same reasoning B also works
lyric obsidian
!nosols
final saddleBOT
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.
pliant shore
so by process of elimination
lofty willow
I'm assuming you're limited to 0.1-0.9?
devout breach
this is the 3rd problem in the same test
silent cove
you double it and it'll give you a a.b form decimal number
devout breach
how can the first one be that hard
silent cove
since last digit is 5, it'll get to 0
magic coyote
how can the first one be that hard
It's not it's just a bit of thought is all
pliant shore
I'm assuming you're limited to 0.1-0.9?
you could also have .0 I think, but it doesn't matter for this q
silent cove
so 0.5*8.7
lyric obsidian
how can the first one be that hard
No, it's a little clever
lofty willow
you could also have .0 I think, but it doesn't matter for this q
Its stated that each box represents 1 digit
silent cove
so yeah i just eliminated the options
lofty willow
So i assumed that its 0.1-0.9 or rather n.0 - n.9
pliant shore
So i assumed that its 0.1-0.9 or rather n.0 - n.9
yeah, n.0 to n.9
lofty willow
If thats the case, then the smallest you'd get by any decimal multiplication by 1 dp with 0.5 would theoretically result in only 2 dp.
So by hypothesis it would only be we'll you can figure that out yourself:")
devout breach
a.5 x b.c = 1/100 (100ab + 10(ac + 5b) + 5c)
pliant shore
a.5 x b.c = 1/100 (100ab + 10(ac + 5b) + 5c)
you're overcomplicating this dude
pliant shore
it worked with 4 * 2 = 8 and 4.8 * 2 = 9.6
.
devout breach
ab + 1/10 (ac+5b) + 1/20 c
wait
wait wait wait ah
since a, b and c are integers there cannot be 3 decimal places since the smallest thing must be 1/20
rancid zenith
you're overcomplicating this dude
How else will you prove option c isnt possible?
lofty willow
this is the 3rd problem in the same test
I think this is quite straight forward?
stuck juniper
pliant shore
How else will you prove option c isnt possible?
by the multiplying by 2 reasoning, whenever you multiply 2 by a decimal with last digit 5, you remove 1 decimal place
devout breach
this is the 3rd problem in the same test
i know how to solve this
but problem 1 is ridiculously hard compared to no.3
pliant shore
so reversing the reasoning, when you multiply by 0.5, you add 1 more decimal place at maximum
rancid zenith
ab + 1/10 (ac+5b) + 1/20 c
Well yes, thats what this also says though?
You can get atmost 5/100 precision
pliant shore
but you are only multiplying n.5 by a 1-decimal place number
devout breach
i found something
3.5 x 3.5 = 12.25
3.5 x 3.6 = 12.6
3.5 x 3.7 = 12.95
so whenever c in a.5 x b.c is odd there is 1 more decimal place
pliant shore
mhm!
devout breach
wow this one is brutal
pliant shore
it is creative but you don't actually have to generalise it all the way
you could just go from 4.35 * 2 = 8.7
so 8.7 * 0.5 = 4.35
devout breach
ah
pliant shore
and then process of elimination gives C
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spare moth
can someone tell me how to do this... what am i supposed to do about the modulus
lyric obsidian
can someone tell me how to do this... what am i supposed to do about the modulus
Find when sine is negative and flip the function in those intervals
final saddleBOT
stuck juniper
can someone tell me how to do this... what am i supposed to do about the modulus
Just plot graph for |sinx|
final saddleBOT
Available help channel!
final saddleBOT
bitter spindle
,rotate
soft zealotBOT
bitter spindle
Sorry I don’t get it
tired walrus
What does the Trident means
it means "belongs to"
it's a symbol from set theory
they're saying that 3x-2 is a natural number
bitter spindle
does natural number means positive integer ?
tired walrus
yes
bitter spindle
Hello!
tired walrus
the constant term
the term which is just a number with no x attached
or in other words, the x^0 term
bitter spindle
Ohhh
I think I kinda get it thank you
Bye bye! Thanks for helping
.closw
.close
final saddleBOT
bitter spindle
.reopen
final saddleBOT
✅
bitter spindle
or in other words, the x^0 term
Sorry I don’t get it again
😭
Term independent of X
tired walrus
do you want:
A) a re-explanation of what "term independent of x" means along with an example in a different expansion?
or
B) a strategy for how to do this particular question
stuck juniper
tired walrus
why are you thanking me so early when i haven't even done or said anything
anyway
basic example
can you expand $\paren{x + \frac{1}{x}}^2$ in full for me please
soft zealotBOT
Ann
tired walrus
(it's not that long)
bitter spindle
x^2 + 1/x^2 + 2
magic coyote
x^2 + 1/x^2 + 2
Correct
tired walrus
yep
so you see that lone 2?
the one at the very end
that's the term independent of x in this.
bitter spindle
Yes
Ohhh
OHHH
OKAY
I GET IT NOW
OK!!!!
THANK YOU SO MUCH!!!!
🙏
have a gooood day!!!
.close
bitter spindle
final saddleBOT
bitter spindle
hello ! What is the name for this vector thingie? So I can search on google and see how it works
And is there a name for the trick like breaking the 10 to 6 and 4?
tired walrus
hello ! What is the name for this vector thingie? So I can search on google and ...
that's not a vector in this context; it's a notation for the binomial coefficient
bitter spindle
ohh
tired walrus
also known as "n choose k" or "n choose r" and sometimes written with a big C and the numbers orbiting around it in some position
(i've seen at least 4 different ones)
bitter spindle
um
Okay!
How about the trick breaking it into 6 and 4?
Is that a better way of pascal triangle
tired walrus
you want the term independent of x so you want the exponent on x to work out to 0
bitter spindle
Yes
tired walrus
the terms will all look like $\binom{10}{k} (x^{1/3})^{10-k} (-x^{-1/2})^k$
soft zealotBOT
Ann
tired walrus
your job then becomes to simplify that thing with the x's until you've got x^(a single power)
and then equate that power to 0 (because independent of x! if you were looking for the x^7 term or something you'd set it equal to 7 accordingly) and solve for k
bitter spindle
the terms will all look like $\binom{10}{k} (x^{1/3})^{10-k} (-x^{-1/2})^k$
Does the term have to be like that?
So if I Wnana use this trick I need to arrange the whole equation into that
( X^(1/3) ) ( -x^(-1/2))?
tired walrus
Does the term have to be like that?
well the form of it comes from the binomial theorem so yes
bitter spindle
ohhh
Thats sounds complicated
But efficient trick
Thank you so much for teaching me how the trick works!
Thats very cool
Have a great day !!
.close
final saddleBOT
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Available help channel!
covert wolf
c) Identify an interval where the graph of g is increasing and concave down.
The graph is g(x) = 10sec(x) + 1
final saddleBOT
c) Identify an interval where the graph of g is increasing and concave down.
The...
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
3
magic coyote
Show your work
covert wolf
tbh i havent even done any work ive just been looking at a graph and i decided the the answer was (pi/2,pi)
covert wolf
tbh not really
magic coyote
tbh not really
Derivative and double derivative and equate to 0, does that sound familiar?
magic coyote
anyone can see me?
!occupied
final saddleBOT
anyone can see me?
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
covert wolf
Derivative and double derivative and equate to 0, does that sound familiar?
yes but i may have forgot it
magic coyote
Study your notes then
covert wolf
ok
noble pawn
Can someone explain to me why tf this works 😭
final saddleBOT
Can someone explain to me why tf this works 😭
tiny gorge
what does "x times" mean if x is not an integer?
spare gate
can anyone suggest me how to learn to solve series questions
sullen gale
can anyone suggest me how to learn to solve series questions
!occupied
final saddleBOT
can anyone suggest me how to learn to solve series questions
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
soft zealotBOT
k
terse crypt
[ \frac{\dd}{\dd x}x^2 = 1 \cdot \sum_{k=0}^{x} 1 + \left( \frac{\dd}{\dd x} \sum_{k=0}^{x} 1 \right) x]
soft zealotBOT
k
terse crypt
[ \frac{\dd}{\dd x} = 1 \cdot x + \left( \frac{\dd}{\dd x} x \right)x = x + x = 2x]
soft zealotBOT
k
formal trail
that's still sideskirting the actual problem of defining what it means for the upper bound of a sum to not be an integer
final saddleBOT
@noble pawn Has your question been resolved?
ripe jewel
that's still sideskirting the actual problem of defining what it means for the u...
i would say it means you're summing for all integers k satisfying 0 <= k <= x
i think that's how Knuth defines expressions like these in TAOCP
formal trail
if that's the definition then it's not true for fractional $x$ that $x^2 = \sum_{k = 1}^x x$
soft zealotBOT
cloud
timber plume
How is there a 3X that is common
final saddleBOT
How is there a 3X that is common
tired walrus
things played out that way
there's not really a deeper reason to it
the first term is 3 * (x^2+5)^2 * 2__x__ * (x^3-1)^4
and the second is (x^2+5)^3 * 4 * (x^3-1)^3 * 3x^2
does that answer your question @timber plume
tired walrus
ok you can pull out only 3 if you so desire
and you'll find that the brackets will have a common factor of x inside
which can also be pulled out
timber plume
Yeah then If it does, I could factorise it further?
final saddleBOT
@timber plume Has your question been resolved?
soft heart
i found the total fail which is 62..
but how 38% is passed in both subjects, instead of it being a total pass
final saddleBOT
i found the total fail which is 62..
but how 38% is passed in both subjects, ins...
proven venture
62% failed at least one exam. then the remaining 38% must have passed both exams (failed 0 exams)
final saddleBOT
@soft heart Has your question been resolved?
soft heart
it is easy to understand..ik...but idk why my brain not clicking
wouldn't it be 38% passed at least one exam
.close
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Available help channel!
raw merlin
Let n be a positive integer. Let k be a point on the number line that represents a real number such that the number of factors left of k to 1 is equal to the number of factors right of k to n. Let p be a factor of n such that p < k. Prove that n > pk.
final saddleBOT
Let n be a positive integer. Let k be a point on the number line that represents...
tired walrus
the number of factors left of k to 1
what does this mean
raw merlin
The number of factors less than k, starting from 1.
tired walrus
the number of whose factors?
raw merlin
n
raw merlin
!status quo
final saddleBOT
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
ok, consider this somewhat abstract phrasing:
let D be the set of all divisors of n, L be the set of all divisors of n which are less than k and H the set of those greater than k. (standing for "divisor", "low" and "high" resp)
and look also at the function f: D -> D given by f(d) = n/d
my claim is that this is a bijection and f(L) = H
raw merlin
Okay. Let me think.
I dont know what you mean by bijection.
bi-jection
bi-jec-tion
tired walrus
I dont know what you mean by bijection.
a function which is both one-to-one and onto
raw merlin
I have thought about something.
Let me type.
root(n) is a valid choice for k.
the greatest factor less than root(n) < root(n), k < the smallest factor greater than root (n)
Clearly, the smallest factor greater than root(n) <=n/p > root(n)
Add [root(n), k < the smallest factor greater than root (n)] in the above inequality.
root(n), k < the smallest factor greater than root (n) <=n/p > root(n)
k < something <= n/p > root(n)
=> k < n/p
So n > kp
Is this correct
raw merlin
I did not let k be root(n)
k, root(n) < something implies:
k < that something
and
root(n) < that something
is my proof correct
rancid zenith
It is true that k=sqrt(n) is a valid choice for k, but you must prove for every valid k
raw merlin
I started from this inequality
the greatest factor less than root(n) < root(n), k < the smallest factor greater than root (n)
rancid zenith
Yes
raw merlin
whats the problem
rancid zenith
No problem here, but your notation is a bit diffult to read
Im assuming you are saying
greatest factor lesser than sqrt(n) < k < smallest factor bigger than sqrt(n)
That is the correct bound for k
rancid zenith
No, you said
the greatest factor less than sqrt(n) < sqrt(n), k < the smallest factor greater than sqrt (n)
raw merlin
k, root(n) < something implies:
k < that something
and
root(n) < that something
rancid zenith
Ok, so thats your notation