#help-39
pearl pondBOT
oblique reef
hello what would be the answer to this?
I don't have an answer key
basically what i did is x^2 = 2^5 * 3^3 * k
in which case x has to have at least 2^3 and 3^2
in which case the answers would be A), B), and D)
robust flax
That seems good to me
pearl pondBOT
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pearl sandal
im not entirely sure what this question is asking but ive given a attempt
could someone explain what im supposed to be doing?
pearl sandal
also when i say x is any number between 0 and 3/2 i meant delta
pearl pondBOT
@pearl sandal Has your question been resolved?
pearl pondBOT
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pearl pondBOT
pearl pondBOT
@jolly sleet Has your question been resolved?
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neat mountain
C=0.0005x³-0.7x²-30x+3000
determine
(i)avg.cost
(ii)margin cost
(iii)value of x at which mc=ac
neat mountain
is this question even doable/correct? ☠️
avg cost and marginal cost is solvable atm
idk tf to do with (iii)
ping me if anybody wannn help
nevermind imma ask from my teacher later
.close
loud geyser
dark ingot
Put x=3 in all the threw and then equate it to 12
3a+3=12
9+3b+1=12
umbral marsh
pearl pondBOT
umbral marsh
I'm not very familiar with these types of problems when there is no c value
I attempted, but was wrong
lone kernel
When there's no c value you just have c=0
dark ingot
pearl pondBOT
jolly parrotBOT
Hack With Techno Boy
plucky oriole
@dark ingot
sonic vault
hi
sonic vault
can I get some help
west sapphire
sonic vault
<:JustAsk:371662163489718273>
question 33
I solved it got the x two values
not sure what to do now you know
how do I know what it will be @west sapphire
toxic lichen
I solved it got the x two values
you mean you solved the equation corresponding to it?
ie |3x-5|=13
@sonic vault
sonic vault
you mean you solved the *equation* corresponding to it?
yes
I know how to solve it
but I odnt know the interval notation
toxic lichen
well let's see
toxic lichen
let's try a simpler equation to see if your memory gets jogged
sonic vault
ok
toxic lichen
|x| >= 10
can you write the solution set of this for me
in interval notation of course
sonic vault
can you write the solution set of this for me
hmm
let me see
no clue
I dont know how it works to be honest
toxic lichen
you don't know how interval notation works at all?
toxic lichen
why did you not say so in the beginning
pearl pondBOT
@sonic vault Has your question been resolved?
midnight haven
Hey, i have
E = {2;{4;5};4}
And it say {4;5} ⊆ E is false
Can you explain why ?
midnight haven
{4;5} has 2 elements
I thought that with the "{}" it only made 1 element
vestal tapir
it is one element
midnight haven
Okay
vestal tapir
it has 2, inside
vestal tapir
it has 1 element
midnight haven
Because there is 2 times {} ?
vestal tapir
yes
velvet meadow
pearl pondBOT
velvet meadow
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pearl pondBOT
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atomic ferry
does parameter p means p=t
pearl pondBOT
does parameter p means p=t
magic hornet
what are t and p?
atomic ferry
lone kernel
I think this is saying to plug in t=p
atomic ferry
can p be taken as a constant when i differentiate
lone kernel
But that is bad notation and the question writer should feel ashamed
lone kernel
Then Cambridge should feel ashamed
atomic ferry
can p be taken as a constant when i differentiate
can?
lone kernel
I don't even think p shows up when you're differentiating
atomic ferry
am i not suppose to sub t=p
lone kernel
Eventually sure
lone kernel
Yes, but if p is showing up when you're computing the derivative then you're doing something wrong
atomic ferry
so i only sub p at the last step?
lone kernel
I mean what do you think
atomic ferry
porbarly
lone kernel
You're finding the derivative at a point
atomic ferry
probarly
lone kernel
What does that mean
lone kernel
It's like if you were asked to compute the derivative of x^2 at x=c
You'd first differentiate and then plug in x=c
Giving you 2c
atomic ferry
okay
lone kernel
R is the real numbers
flat kayak
pearl pondBOT
magic hornet
,rotate
jolly parrotBOT
flat kayak
after finding the limit i need to determine if its continuous
stark summit
why do you think that limit approaches 0?
flat kayak
idk i worked through it and got 0
stark summit
kay now to determine if its continuous what do i do
you might start by looking at the domain of the function
flat kayak
idk
stark summit
idk
you may need to break up your function by it contents
so if we look at the function
we see a cos, a sin and a tan
then see the domains of their functions
like what is the domain of cos(x)?
flat kayak
all real numbers i think
flat kayak
all real numbers
stark summit
and tan(x)?
stark summit
it has a lot of asynptotes
yep does that mean we can add all values in?
stark summit
so its not continuous cus there would be an asymptote at 2pi?
might relook at the values that cannot be added in tan(x)
flat kayak
idk
stark summit
so its not continuous cus there would be an asymptote at 2pi?
btw do we need to find if the function is continuous in all values or continuous in 2pi
because i can't see the original quesiton
stark summit
so is 2pi part of the domain of tan(x)?
flat kayak
yes
stark summit
so is tan(x) continuous at 2pi?
flat kayak
yes
stark summit
so is sin(9pi/2cos(tan(x))) continuous at 2pi?
flat kayak
yes
stark summit
well there we go
flat kayak
so is that correct
if it wasnt continuous would it not work
so say it was approaching pi/2 isntead
since tanpi/2 is undefined
it wouldnt be continuous?
stark summit
yep
flat kayak
does not exist
flat kayak
i dont think so
i think thats it for now
so if the limit doesnt exist at the value its approaching
its automatically gonna be continuous?
and if it doesnt exist, its automatically not continuous
ok yea i remember from the lecture
thank you
stark summit
np
pearl pondBOT
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prime bramble
pearl pondBOT
prime bramble
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pearl pondBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'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
So I know that I need a number a+bi that squares to get 5 + 12i
I know that 5 = a^2 - b^2 and 12 = 2ab
but I'm having a hard time actually solving this system because I'm getting nested roots, which I'm not sure is right
foggy mason
a = 6/b
(6/b)² - b² = 5
36/b² - b² = 5
36 - b⁴ = 5b² (if b≠0)
b⁴ + 5b² -36 = 0
its solvable I guess
prime bramble
a = 6/b
(6/b)² - b² = 5
36/b² - b² = 5
36 - b⁴ = 5b² (if b≠0)
b⁴ + 5b² -36 = 0...
yeah but this is a huge pain
I was wondering if there was a nicer way
jolly parrotBOT
foggy mason
b² = -9 or b² = 4
b cant be complex
thereforce b = 2 or b= -2
then a = 6/b
a = -3 or a = 3
its done
prime bramble
b² = -9 or b² = 4
b cant be complex
thereforce b = 2 or b= -2
ah, I forgot a,b have to be real
fml
thanks
.close
pearl pondBOT
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twin oar
what is considered a distinct vector?
pearl pondBOT
what is considered a distinct vector?
timid spindle
Context?
twin oar
it wont let me send a screenshot
ill copy and paste the text
it also wont let me copy anything new..
ima restart..
now i need to fill in 2fa
woohoo
@timid spindle
timid spindle
Right okay
So it's not talking about the vectors having a property called "distinct"
Here it means they're "distinct from eachother" I.e. different from eachother
twin oar
u v and w btw are in multiples of each other
well not rlly they all just simplify to (1,1,1,1)
twin oar
Here it means they're "distinct from eachother" I.e. different from eachother
how would you tell if theyre different?
timid spindle
Well you see if you can find two different vectors
twin oar
i can find two different vectors that are simply in multiple
but i can also find an infinite amount of vectors that are in multiple
r4 means rank 4 right?
timid spindle
If you can find two that are different then you're done
twin oar
If you can find two that are different then you're done
not rlly cuz it says exactly two lol
timid spindle
Eh question is vague, taking two vectors that satisfy the property can be interpreted as having exactly two imo
twin oar
maybe ill do that and also say that you can have R(vector) amount where R is all real numbers
technically couldnt you also do complex numbers?
oh no cuz then it wouldnt be orthogonal
timid spindle
If it means "are there two vectors and no more with this property" then yeah what you're saying applies
twin oar
my teacher made a point to say in class that "mathematicians are cheap with words"
so i think it means what it means
btw i have a question
is there a quicker way to notate matrix modifications?
or a way to show modification to multiple rows at the same time?
or do i just have to deal with the tediousness of that
timid spindle
As opposed to like R1 = R1 + 2R3 etc
twin oar
yeah
timid spindle
Not really
jolly parrotBOT
Joshii
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
twin oar
shoot
timid spindle
\to
timid spindle
Uh but that's pretty much the same in terms of ink and speed lol
jolly parrotBOT
Joshii
twin oar
yeah i know
hopefully my prof wont give anything too long then when it comes to testing lol
timid spindle
You can probs get away with just like 5R2 if you're just scaling a row
But honestly it's not too bad
twin oar
yeah just gotta get used to it
its mostly my handwriting thats an issue and i take up unecessary space with other stuff before i think to write down the notation
timid spindle
Just encourages you to be efficient with your row ops so that you have to do less of them lmao
twin oar
yeah the second time i did it i got it down to 10 instead of 12
for that question
part a
pearl pondBOT
Available help channel!
pearl pondBOT
keen marlin
Anyone know where to begin here?
I know that (0,0) is a point since z is undefined there...
pearl pondBOT
@keen marlin Has your question been resolved?
keen marlin
<@&286206848099549185>
cunning comet
I know that (0,0) is a point since z is undefined there...
thats all.
keen marlin
How can I phrase that there are no other points where the function is discontinous?
or in other words, prove that (0,0) is the only point...
cunning comet
hmmm, is this really necessary? sum ,products, ... of continous functions are continuous except quotients where the denominator has to be <> 0. so you have to look for this. only imho.
keen marlin
hmmm, is this really necessary? sum ,products, ... of continous functions are co...
i'd like to know for my own knowledge lol
what does <> mean?
cunning comet
not equal
keen marlin
ohh
cunning comet
ok, if you want it this way. 2. (f is defined) is clear, 1 and 3 are limits. so you can use limit rules. limit of sums is sum of limits, .... which is just the same as sum of continous functions is continous ...
keen marlin
is it that the parts that make up the entire function
are continuous
so the enitre funciton is continuous
except at (0,0) ofc since the function is not defined there...
adding 2 continous functions, results in a continous function right?
keen marlin
gotcha
so
(x^2-y^2) is continous
and xy is continous
so the product, xy(x^2-y^2)
is continous?
cunning comet
yes. yes.
keen marlin
hmm
well that leads me to my follow up...
how in the world would I check for this funky looking function?
cunning comet
dont be afraid of the looking of the function. lets do it step by step. start with the y^2/(...) fraction -> denominator must not be 0.
cunning comet
ok, lets go to the second part. its an e^(...) which is normally continous - if the exponent is defined. so look at the exponent only. what is the exponent?
cunning comet
negative does not matter.
keen marlin
so is it e/(whatever is rasied to the power)?
keen marlin
negative does not matter.
oh
cunning comet
e^(...) is defined from minus infinity to + infinity, so the sign does not matter. its a fraction what matters. the fraction has to be defined.
and for the fraction you have only to look at the denominator.
keen marlin
I believe that the denominator
is defined for all values of x and y
is it is (...)^ + (...)^2
cunning comet
is it is (...)^ + (...)^2
look at the first part, the y^2/(...) part. there the denominator is also (..)^2+(,,,)^2 and you have discontinuity.
keen marlin
wait
how is there discontinuity there lol?
oh wait
nvm
the reason the e^(....) denominator is never 0 is because of the -1
wait no
(1,0)
makes that denominator 0
cunning comet
yes, and there is another (x,y)-pair. which makes the denominator 0
keen marlin
So the only points of discontinuity are (0,0, (1,0), and (-1,0) ?
cunning comet
well, at this points the function is not defined. and therefore it cant be continous. but in general (not in this case) you have to be careful. continuity has 3 criteria, and we are talking about the second one (f is defined), but the first is fulfilled at (1,0) and (-1,0).
keen marlin
well
if one of the three fails, then it can't be continuous at that point right?
one of these 3
oh
cunning comet
yes, thats right. but my point is, you could easily make this function continous at (1,0) and (-1,0) with a simple additional definition..
keen marlin
hmm i'm confused
so r we saying the function is not discontinous at (1,0) and (-1,0)?
cunning comet
no. i am saying this function is discontinous at the three points.
but wait a moment
keen marlin
but wait a moment
?
keen marlin
oh
just out of curiousity and learning
Can we make fill out this piecewise function so that for z that removes as many holes as possible by making z equal to the limit value.
cunning comet
its necessary to check always all three conditions of continuilty.
if one condition fails, its discontinuity
keen marlin
so
let's do (-1,0) first
Do we just simply pick a number or would it be another function on it's own?
cunning comet
the e^(...) part is not defined for (-1,0). thats what we haed.
but in the modified version we have defined f(-1,0) as 0, so it is defined.
so we have to check the two limit criteria: does the limit exist and is it equal to the function value.
so what happens with the exponent when you have something going to x = -1, y = 0
keen marlin
the exponent wants to be 0
cunning comet
no, the denominator in the exponent will go to 0. so the fraction will go to infinity, so the whole exponent will go to -infinity.
keen marlin
ah yeah
cunning comet
and e^(-inifinity) will be 0.
keen marlin
gotcha
so the function value will be 0
no
yes
it will be
no no
wait
yes
but isn't finding the limit for this so tedious?
cunning comet
so the limit exists and equals the (defined) funtion value.
keen marlin
how do we know the limit is 0?
oh yeah nvm
we just did that
but the function value is still undefined
so we need to add that to the peiecewise functin
so what exactly r we adding?
cunning comet
so we need to add that to the peiecewise functin
that is what i did. look at my picture.
cunning comet
wait a moment.
cunning comet
you do not need to do this.
i just wanted to explain ...
.... that you need to be careful.
keen marlin
Yeah u were just explained that we could make it continuous by creating that peiceswise function?
Right?
cunning comet
exactly.
keen marlin
There are no holes anymore in that piece wise right?
cunning comet
(0,0) is still here.
cunning comet
i dont think so. but i am not sure. we would have to check if the limit exists.
cunning comet
well, we can try it.
keen marlin
we know that this function is discontinous at (0,0), (1,0) and (-1,0)
I need to now find the limits of the function
at these points
I have to find the limit as (x,y) approaches these points
and I am given this fact to use:
Isn't this very rigourous for all three? or are a couple of them straightforward...
idk how much multivariable calc u know or not...
cunning comet
we used this fact, and we calculated the limits for (-1,0) and for (1,0) . in my "modified" function.
keen marlin
hmm
but we can't perform direct substitution to solve these limits...
so would we have to use squeeze theorem?
i'd like to know how we would write out the full entire steps to solving the limits at these points...
keen marlin
where do we use this fact?
cunning comet
its the same argument. if g(x) goes to 0, |g(x)] goes to 0, so the fraction goes to infinity, so the whole exponent goes to - infinity, so e^(...) goes to 0.
keen marlin
yes
we have to show the enitre multivariable limit
like (x,y) -> (1,0)
so I can't split it up...
wait...
cunning comet
if (x,y)->(1,0) then ((x^2-1)^2+y^2)->0, then 1/(...) -> infinity then -1/(...) -> -infinity then e^(...) -> 0
keen marlin
r we allowed to split the limit into 2?
because we can't ignore
the y^2/(x^2+y^2) ...
If both limits exist, then the limit of the sum is the sum of the limits. And if one of the limits exists but the other doesn't, the limit of the sum doesn't exist.
is that true?
cunning comet
the y^2/(x^2+y^2) ...
this will end in a 0 / (1 +0) = 0
keen marlin
yes
cunning comet
If both limits exist, then the limit of the sum is the sum of the limits. And if...
yes
keen marlin
so I can show that the first part if (x,y) -> (0,0) is 0
and then with another limit show that
keen marlin
if (x,y)->(1,0) then ((x^2-1)^2+y^2)->0, then 1/(...) -> infinity then -1/(...) ...
this.
since both seperate limits r 0
the sum (or subtraction) of the original limit
is 0?
cunning comet
so I can show that the first part if (x,y) -> (0,0) is 0
i am not sure if this holds. i have to think about this.
cunning comet
so I can show that the first part if (x,y) -> (0,0) is 0
ok, woframalpha says its true.
keen marlin
ok, woframalpha says its true.
well, we can just sub the values in no? and since it is defined it works...?
cunning comet
if (x,y)->(0,0) then y^2/(x^2+y^2) -> 0/(0+0) = 0/0 which is not defined and could be (as limit) everything.
so you have to argue in another way.
keen marlin
well
we r only talking about (x,y) -> (1,0) right now
so for (1,0) we can just plug them in and it gives 0, do u agree?
cunning comet
we r only talking about (x,y) -> (1,0) right now
do we? you wrote about (x,y)->(0,0) so i thought we are talking about this (see: #help-39 message)
cunning comet
no problem. makes it easier of course. yes for (1,0) it is all fine.
keen marlin
for (1,0) we can plug in for that one part and then the other part we check seperate and use the given fact
so the function as it approached (1,0) is 0
that we agree on?
cunning comet
yes, that was my argument how we could make the function continous at the two points.
keen marlin
ok
so now we have to check the limit of the function at (-1,0)
is the limit and the steps not the exact same as is was for (x,y) -> (1,0)?
cunning comet
it is just the same, as we have x^2, so 1^2 and (-1)^2 is the same.
keen marlin
so for both (1,0) and (-1,0) the limit is 0?
cunning comet
yes.
keen marlin
ok so now we just have
(x,y) -> (0,0) left to check
wolfram says this limit does not exists
usually to show that a limit does not exist
we can find two equations that go through (0,0) for example y = 0 and y = x
and show that plugging each of the in give 2 seperate numbers (limits) so the limit DNE
Do u see how we could potentially do that here?
cunning comet
hmmm, i am not sure. wolfram says the limit of the first part exists and is 0. when i look at the e^(...) part, this would be e^(-1), so why should the limit not exist?
keen marlin
hmm
let me write it down
one sec...
wait
if we consider
(y=0)
and
(x=0)
we get two different limits tho...
no?
cunning comet
sorry, obviously i had a mistake. now my wolfram also says "does not exist"
keen marlin
yeah
are we allowed to approach this limit along y = 0 and x = 0?
i'm not too sure about the rule
does it matter if both have to be y = ... or x = ...
or can we do y = 0 and x =0 ?
cunning comet
i am not sure, what you want to do. do you want to prove that the limit does not exist?
keen marlin
yes
cunning comet
so you only need two different sequences.
keen marlin
yes
y = 0
yields a different limit
from x = 0
so that proves it
but i just don't know if i can do that
do I have to use like y = 0 and another y = ...
or can one be x = ...
and one be y = ...
cunning comet
for x = 0, the first part becomes a 1, so the limit would be 1 in this case.
keen marlin
i'm talking about the limit as (x,y) -> (0,0)
cunning comet
yes, i do also. for example fix x = 0, and y = 1/n, so the first part y^2/(y^2+x^2) will be y^2/y^2= 1, so the limit is 1. for this sequence.
keen marlin
but can't we just say to show that the limit does not exist, we must find two different directions of approach to (0,0) which have different limits..
and use y = 0 and x = 0
cunning comet
for the second direction use, y= 0 and x = 1/n
keen marlin
for y = 0 we get lim to be −e ^−1
for x = 0 we get lim to be 1 −e ^−1
different limits
therefore, the limit DNE
isn't that easier?
cunning comet
if its fine for you, its ok. you have to feel comfortable
cunning comet
for y = 0 we get lim to be −e ^−1
i didnt get this argument.
if y=0 then y^2/(x^2+y^2) will give you 1/x^2 which will be infinity for x -> 0
keen marlin
Lim(x->0) [ (0^2 / (x^2 + 0^2)) - e^(-1 / ((x^2 - 1)^2 + 0^2)) ]
= Lim(x->0) [ 0 - e^(-1 / ((x^2 - 1)^2)) ]
= e^(-1)
cunning comet
we are talking about different things.
cunning comet
you want to show that the limit for the function f does not exist, if (x,y) -> (0,0). the limit for the e^(..:) exists (IMHO) and is 1/e. so you have to show that the limit for the first part y^2/(x^2+y^2) does not exist.
keen marlin
The limit exists only if the value of the limit along every direction that leads to (0,0) is same.
if we approach along y =0
and x = 0
we get different limits
therefore, it DNE
cunning comet
show.
keen marlin
ok let me write it down
keen marlin
first: i dont believe this part:
well it is true lol
cunning comet
ignore the e ⁽..) part, this limit exists.
keen marlin
u can check on wolfram too
cunning comet
well, then all is done, isnt it?
keen marlin
if you could show that the limit as x -> 0 of z(x, 0) is different from the limit as y -> 0 of z(0, y), or that either limit fails to exist, then you will have proven that the limit as (x, y) -> 0 of z(x, y) doesn't exist
that is what we have done
keen marlin
well, then all is done, isnt it?
yep
thanks a lot for taking the time to keep up with me lol
truly appreciate the help!
pearl pondBOT
pearl pondBOT
@smoky crystal Has your question been resolved?
pearl pondBOT
@smoky crystal Has your question been resolved?
vestal tapir
it's just the third option
you can subtract the answers to a and b
since you have them already
oh wait, that's not the order
@smoky crystalwhat's the answer to a) ?
just do the same thing with 2 Es and then subtract 1/26
there's 12 palces for EE to go
ok, that's probably just wrong
desert oyster
There's couple of ways you can do this
Do you know the number of solutions to a+b+c=10 where a,b,c are all whole numbers
Ok no worries
We can do it the normal way
So try to find all the combinations where atleast 2E are together
If you think of EE combo as a separate letter X
vestal tapir
the simplest way is like, if you take 10 remaining letters, there's 11 places to put stuff around
midnight haven
Any IIT JEE aspirants here
vestal tapir
so 11×10 ways to put E and EE
desert oyster
So try to find all the combinations where atleast 2E are together
Then this will contain cases where words will have ..EX.. and ..XE..
midnight haven
??
desert oyster
So you just need to subtract twice of the total combinations from the part A
Wait hmm
Wait i think i might be double counting, one sec let me check
vestal tapir
ntrtainmnt is 10 letters
.n.t.r.t.a.i.n.m.n.t. has 11 spots
you put E into one, and EE into other, and you got everything
it's the shorteest way
vestal tapir
no
you could, but then you multiply by 2, for E before EE
so you just 11×10 instead
smoky crystal
so you just 11×10 instead
sorry what does 11 * 10 give ?
like what does it mean
how many ways for wat
vestal tapir
yes
smoky crystal
So you just need to subtract twice of the total combinations from the part A
pls @ me ifu ever find out i like to see other ways :)
desert oyster
pls @ me ifu ever find out i like to see other ways :)
I think it's wrong and it requires some changes cos grouping of things is a bit complicated than it seems at first sight XD
But yeah the beggar's method is a very helpful method to solve such questions
The one which you guys used
pearl pondBOT
pearl pondBOT
Available help channel!
broken burrow
Which one is the correct and more accepted sign to use in multiplication?
$ 4 \cdot 5 $
OR
$ 4 \times 5 $
jolly parrotBOT
chlamydia
midnight haven
both work. it depends on your preference and background
notation is often subjective
[
4\cdot 5, 4 \by 5, (4)(5)
]
are all valid
jolly parrotBOT
broken burrow
Okay, I was just wondering that for example on research papers which is the accepted way, but as it turns out it doesnt matter
golden bramble
It also depends on the course honestly
Pre-alg x is ok for multiplication, but for alg it’s sacrilege
broken burrow
Pre-alg x is ok for multiplication, but for alg it’s sacrilege
wdym by sacrilege
warped violet
violation or misuse of what is regarded as sacred.
broken burrow
violation or misuse of what is regarded as sacred.
Ohh gotcha
midnight haven
Pre-alg x is ok for multiplication, but for alg it’s sacrilege
You'd be mistaken. There are a few graduate students I have met that use × for multiplication. Granted, they are in the minority.
warped violet
Have never met anyone who uses x
timid spindle
Okay, I was just wondering that for example on research papers which is the acce...
A research paper would in nearly all scenarios not use x
midnight haven
Use whatever you want. Just be consistent and clear. Don't try to mix-and-match stuff
broken burrow
A research paper would in nearly all scenarios not use x
Gotcha
Yeah it is weird, because I head that in the US they are using the \cdot but in Ireland for example that would mean the decimal point so that's why we use the x for multiplication.
vestal tapir
if in ireland that means the decimal point, then it makes sense
no reason to doubt it
broken burrow
btw, what are you guys using for decimal point then? if you'd stick with the dot sign.
golden bramble
I use the dot ala 3.99
Some cultures swap around the comma and the dot
So 3,999.75 in one place may be 3.999,75 in another
broken burrow
I use the dot ala 3.99
Yeah It can be confusing especially if you are international cause I'm originally from Hungary and we've done it the exact opposite way that we are doing here in Ireland now 😄
timid spindle
Yeah it is weird, because I head that in the US they are using the \cdot but in ...
Fwiw, the two dots have different positions, the multiplication dot is in the middle and the decimal point dot is at the bottom
Im from the UK and we frequently use cdot and the decimal point together with no confusion
broken burrow
Fwiw, the two dots have different positions, the multiplication dot is in the mi...
For us in Ireland we use the decimal point in the middle 😄 but yeah it would make sense to put it at the bottom IDK why they are not doing it, but that's why we are using x for multiplication
golden bramble
In your case then, probably best to use parentheses
timid spindle
I honestly don't believe that Ireland puts the dot above the line
wraith river
For c I got 3234cm^2 but apparently thats not correct, can anyone explain why
pearl pondBOT
For c I got 3234cm^2 but apparently thats not correct, can anyone explain why
@wraith river Has your question been resolved?
wraith river
<@&286206848099549185>
dense goblet
shadow tusk
dense goblet
and you already have a channel
dense goblet
For c I got 3234cm^2 but apparently thats not correct, can anyone explain why
3234 seems correct but i'll let someone else confirm
wraith river
3234 seems correct but i'll let someone else confirm
alright
i think for tray maybe the top is open
dense goblet
oh
dense goblet
yeah if it's a tray then the external area is different
have you checked the answer
dense goblet
yeah that's what i got
wraith river
.close
vestal tapir
@smoky crystalyou're assuming the last person is the one who repeats a choice
you could multiply by 4
no
you multiply by 4 again, in case he doesn't repeat at all
it's not a thing, the 5th person could pick something new, and there would still be 4 total
you;re accounting for picks like 32142, 51433
you're not accounting for 12135
in general, 4^5 − 3^5 would be exactly 4 out of 4
,calc (4^5 - 3^5)*5
jolly parrotBOT
Result:
3905
hm
and that's your way
at least one is wrong
i'm confused lemme think
okay yeah, you can't just multiply by 4 for 4 people who could repeat a previous order
you would choose 2 people who order the same thing
because it doesn't make sense
just i thought, you;re saying the last person repeats, but other people could repeat, but the first person can't
so times 4
it just doesn't make sense overall
hmm
that feels right, it's what you;re doing, but we're not multiplying by 4 at all
we decided 2 people who get the same order, so we just do 4! and that's it
i need to think
2 people have to choose the same order
and 4 items will be ordered
and there's 4! ways to assign these 4 items
yes
5*4/2 = 5c2
you did ×4 but there's also 5 spots where this person can go
and divide by 2 because you will overcount
by placing them before or after the person who they "copied"
it's not how you should think, you can just do 5c2 from the start
vestal tapir
you pick two special people, and you choose 4 items, so 5×4×3×2
vestal tapir
it can be writtten as 5c4 × 4!
yes
and then you have 4 "groups" to asign 4 items out of 5
like 4 "people" have 4 meals
and it doesn;t matter anymore that there's 5 people
you just have to pick 4 times
you could start with the two guys, or end with two guys, or go in order and pick for the two guys, when you hit one of them
if there was no constraint
we would go 5×4×3×2×1
it's just, at some point one of the guys will already have picked a meal, by copying some other guy before
meaning we simply skip him
right, assuming they are all different
we want to assign 4 different meals
yes, or 5c4 × 4!
that's the same thing
no, 5×4...
i mean yeah
i'm just saying you can assign 4 out of 5 in one step, 5p4 some would say
yes, and at some random point rthere's a person with 1 option
in the middle
because we already said he's paired with someone else
doesn't matter when or whom, it's 1 option for them, we can ignore them entirely
so it's like
first person has 5 meal options
2nd perosn has 4
3rd and 5th person has 3
4th person has 2
no matter what happens it's the same calculation, what changes is what line is changed to include some other person later
well sure
in some sense
in ourt calcualtion they have 1
first person has 5 meal options
2nd and 3rd perosn has 4
4th person has 3
5th person has 2
we won't
they don;t have 5 options
they don;t exist, because we ensured they are accounted for when we grouped them with someone else
we don;t know their number
all we know is we have to choose 4 times
to assign 4 different items
uh
i meant 4! instead of 5!
like suppose we want 3 people to get the same meal
so 3+1+1, three different meals
try to do the same logic
yeah, and we do 5c3
to account for the pairings we can do
5c3 × 5 × 4 × 3
or 5c3 × 5c3 × 3!
this should be the answer
afraid not 😦
yes, we want to to "tag 2 people with specialness" out of 5
yes
i think i didn't realise you have the problem with that
because it's hard to understand why would it divide 5 people into 4 groups like where's the 4 in 5c2
but then you;re saying you don;t understand why 5c2 chooses 2 people
and like, what else does it do
i don;t see what tree
from the start you were doing things i understand
you assigned 4 meals to 4 people, and you did 5c4 before that
and then there was a remaining 5th person who copied one of the other 4 so you muliplied by 4
i basically just corrected that, the 5th person is not necessarily the one who copies
but then you can't say it's times 5: the first person is definitely not the one who copies
and you can't say it's times 4, because actually only the last person can copy one of 4 people
the second person can copy one etc.
so it's like, instead of 4, you do 1+2+3+4
you simlultaneously choosing the person who copies, and the person who is copied
which is obviously done by choosing 2 people
and that's great
yeah
oh it's mine now
.close
midnight haven
pearl pondBOT
midnight haven
I have defined the addition and scalar multiplication for elements of V
I have no clue how x1 + x2 + x3 = 0 make a connection to whether the set V is a vector space over F.
If anyone is so kind to respond please baby-step me to the solution
What I know of a vector space is that the operation of addition and multiplication of elements in V create other elements within V and that satisfy the 8 axioma's.
pearl pondBOT
@midnight haven Has your question been resolved?
midnight haven
<@&286206848099549185> 🙂
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
@midnight haven Has your question been resolved?
dusk dagger
Are you sure your definition is right? How would you multiply two vectors in R^2?
midnight haven
Mutliplying two vectors does not provide another element within V. It will only provide a scalar.
Scalars are no elements of R^2
But more importantly, there is no such required axioma that must be satisfied that includes the multiplication of two vectors
dense trench
hi i need hlep on this homework
pearl pondBOT
hi i need hlep on this homework
dense trench
i found the r_u & the r_v and found the magnitude of the cross product. it turns out to be (v^2+u^2)sqrt(8) but I don't know what the bounds of integration would be
pearl pondBOT
@dense trench Has your question been resolved?
pearl pondBOT
@dense trench Has your question been resolved?
midnight haven
pearl pondBOT
midnight haven
How would i do this seeing as its cos^2
merry python
Notice that x = x/2 + x/2
merry python
Both are possible, but LHS would be easier
midnight haven
okay thanks let me try that
merry python
Start with cos(x) and use algebra to get to the RHS
wary bane
!show
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
toxic lichen
can u help me
!occupied
pearl pondBOT
can u help 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).
celest crest
ok can i ping u
toxic lichen
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'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
ok nvm i have to go to sleep now
pearl pondBOT
@midnight haven Has your question been resolved?
broken delta
pearl pondBOT
broken delta
can I get a hint on how I could solve this
midnight haven
can I get a hint on how I could solve this
what do you know about the determinant of matrices with linearly dependent rows
broken delta
what do you know about the determinant of matrices with linearly dependent rows
oh wait this has to do with vectors, my class didnt cover those yet
so nvm
.close
pearl pondBOT
broken delta
wait does it?
pearl pondBOT
Available help channel!
unique widget
I have a Calc I question in regards to intervals of increase and decrease
unique widget
I don’t think I did this right; in the other practice problems I’ve been doing I’ll get an easily factored polynomial that spits out two x= values, this one I got stuck with -3 and maybe zero from pulling out the x^2 but it didn’t give me a min and max
Thanks in advance 🙏🏻🙏🏻
midnight reef
I don't see problems in your work
unique widget
So then this just wouldn’t have a local maxima?
And would the x values be x=0 and x=-3 since I pulled out the x^2 from the derivative?
midnight reef
So then this just wouldn’t have a local maxima?
it does not have a relative maximum
midnight reef
And would the x values be x=0 and x=-3 since I pulled out the x^2 from the deriv...
the x values for what?
unique widget
From factoring
I pulled the x^2 from the derivative to split it for favoriting; I would set the x^2 to zero as well, right? Just double checking
midnight reef
They only weird thing I notice is that your zero looks like an empty symbol
unique widget
Oh sorry, I write zeros with a slash through then but I guess I shouldn’t do that moving forward in Calc..
Thanks for mentioning that
Cool, thanks so much for checking my work, I really appreciate it 🙏🏻🙏🏻
.close
fiery drum
I just want to confirm if my answer is right for this question. Basically, I need to find the errors in this proof - and for this question I have found two. I was just wondering if anyone can confirm these for me.
So firstly, the inductive basis is incorrect. P(1) !: 3/2 - 1/1 = 1/(1x2) because by the formula, the RHS should be 1/((n-1) * n). However in this case n - 1 would be 0 and you would have a 0 in the denominator which is not allowed (division by 0).
The next is that for the inductive step, the RHS where 3/2 - 1/n + (1/n - 1/ (n+1)) != 3/2 - 1/(n-1)
pearl pondBOT
I just want to confirm if my answer is right for this question. Basically, I nee...
@fiery drum Has your question been resolved?
fiery drum
<@&286206848099549185>
pearl pondBOT
@fiery drum Has your question been resolved?
dense goblet
The next is that for the inductive step, the RHS where 3/2 - 1/n + (1/n - 1/ (n+1)) != 3/2 - 1/(n-1)
but the working says 3/2-1/(n+1)
pearl pondBOT
@fiery drum Has your question been resolved?
fiery drum
> The next is that for the inductive step, the RHS where 3/2 - 1/n + (1/n - 1/ (...
oh i meant n+1 sorry
dense goblet
other than that, it just looks like n=1 is wrong
fiery drum
other than that, it just looks like n=1 is wrong
so the base case only?
fiery drum
.close
pseudo epoch
Why does one number have to be even and one number have to be odd in question 3.18?
sinful abyss
The question is asking for all pairs of primes whose sum is equal to 61
61 is an odd number
only odd+even=odd
pseudo epoch
Why
sinful abyss
o is an odd number, and $o=2k+1$ (k is an integer)
jolly parrotBOT
Reading
sinful abyss
e is an even number, and $e=2a$ (a is an integer)
jolly parrotBOT
Reading
pseudo epoch
Thanks
pseudo epoch
.close
pearl pondBOT
buoyant stump
coarse dawn
.close
pearl pondBOT
last moth
uh
brazen vector
pal you have the right ask math questions here and get an explanation and a help but you cant be rude to someone helping you
last moth
pal you have the right ask math questions here and get an explanation and a help...
they're banned now anyways
for saying a slur
brazen vector
whats a slur
frosty maple
curse word
brazen vector
english isnt my main language so idk
last moth
like, a word that's used to insult some race/ethnicity/other
coarse dawn
yea then he shouldnt be here if he has this mentality
And the fact he opened a new channel to get help because he didn't understand what I was asking
brazen vector
And the fact he opened a new channel to get help because he didn't understand wh...
yea this was rude
brazen vector
Who becomes rude to people trying to help them
😂
brazen vector
idk man people like this should find a way to get rid of this trash way of thinking
pearl pondBOT
Available help channel!
shy ginkgo
if f(x) = 2x^100 then f'(x) = 200x^99?
pearl pondBOT
if f(x) = 2x^100 then f'(x) = 200x^99?
light helm
no
blissful cloak
if f(x) = 2x^100 then f'(x) = 200x^99?
t is constant with respect to f(x) :)
light helm
or if t and x are related, you'd need chain rule