i am thinking
#help-49
worldly pine
lyric charm
new ratio of A to B in the vessel is 7:9, so 16 parts total and B is 9 of them
worldly pine
new ratio of A to B in the vessel is 7:9, so 16 parts total and B is 9 of them
can you show the solution
lyric charm
i meant $\frac{5}{12} + x = \frac{7}{16}$ not $\frac{5}{12}x = \frac{7}{16}$
grand pondBOT
Ann
lyric charm
also no im not going to show you the solution
did you do step 1 from what i wrote? looks like you did not.
worldly pine
did you do step 1 from what i wrote? looks like you did not.
.
lyric charm
incomplete
the removed portion contains _____ liters of A and _____ liters of B
i think i might need to disappear soon
worldly pine
why
worldly pine
i think i might need to disappear soon
?
worldly pine
out of energy
oh..you are very active on this server...you should be paid for this
lyric charm
and don't want to frustrate myself like this
worldly pine
dont want to repeat every instruction 5 times for you to do it
english is not my native language..i need time to understand it
hushed mauve
is this not a proportion problem
midnight plankBOT
@worldly pine Has your question been resolved?
worldly pine
.close
midnight plankBOT
midnight plankBOT
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lapis pawn
It’s a physics question technically but all math. THE question is : Estimate how many years it takes Mars to orbit the sun
midnight plankBOT
It’s a physics question technically but all math. THE question is : Estimate how...
fair frigate
isnt this techainnly space based
lapis pawn
The only space thing is knowing the days of an orbit
The rest is math
And the days aren’t accurate in this question
If you presume 1.0 is 365 days how would you get Mars if you presume it is 687 days
I guess 687 divided by 365
hushed mauve
you are supposed to estimate
lapis pawn
Which is 1.88 two decimal places
fair frigate
yea i think 1.88 should work no?
hushed mauve
so you could put like 1.8, 1.9, even 2.0
it's an estimate. as long as you state your assumptions, should be good
fair frigate
since be based it off of earth years
hushed mauve
also, i see all the numbers being given to 1dp
maybe, just maybe, stick with that format
unless you have instructions otherwise
lapis pawn
Ok thank you just wanted to make sure
midnight plankBOT
If you are done with this channel, please mark your problem as solved by typing .close
left herald
hello
left herald
gaunt nimbus
progress?
scarlet pond
what r u confused abt
scarlet pond
thats one way to do it
left herald
how would you do it
scarlet pond
i would probably do it that way
left herald
or do you have a more simple way to do so
scarlet pond
that is probably the easiest way
left herald
i get confused with the powers
scarlet pond
or else u have to use quotient rule
left herald
so i get 8w^2 / root w and get stuck there
gaunt nimbus
split fraction -> power rule
scarlet pond
well do you understand how to split it into 3 fractions first?
gaunt nimbus
so i get 8w^2 / root w and get stuck there
${\sqrt{w} = w^{1/2}}$
grand pondBOT
k
gaunt nimbus
remember ur exponential rules 
left herald
like w^n − m?
gaunt nimbus
mhm
scarlet pond
Sorry my disocrd just died on my computer 💀
left herald
so for the first part would it be 8w^3/2?
scarlet pond
U can just cancel top bottom
left herald
if i did 8w^2 - the w^1/2
left herald
U can just cancel top bottom
wdym?
round parcel
so for the first part would it be 8w^3/2?
Yes.
scarlet pond
if i did 8w^2 - the w^1/2
This is right
round parcel
(8w^{\frac32})
grand pondBOT
Chai T. Rex
round parcel
No, you have (\frac6{w^{\frac12}} = 6w^{-\frac12}).
grand pondBOT
Chai T. Rex
round parcel
You can use the power rule on that.
grand pondBOT
Chai T. Rex
left herald
Thats what i figured
round parcel
so then for the -4w/w^1/2 would that be -4^1/2
Yes, but don't forget to type in the w.
round parcel
,rccw
grand pondBOT
left herald
but is that the correct way to pu it
round parcel
Yes, but still a bit of missing stuff.
The line before the last should have +s.
(\frac{8w^2}{w^{\frac12}} + \frac{-4w}{w^{\frac12}} + \frac6{w^{\frac12}})
grand pondBOT
Chai T. Rex
scarlet pond
Second sign is minus
round parcel
It's in the numerator.
scarlet pond
Ah
round parcel
You can also do the - instead of the + there.
left herald
wdym +s?
round parcel
Yours have no plusses between the terms.
left herald
You can also do the - instead of the + there.
where
scarlet pond
U r missing a sign before ur last fraction
round parcel
See the second to last line here.
round parcel
**Chai T. Rex**
See this.
left herald
oh oh yeah i just wrote that for myself
round parcel
Oh, OK.
round parcel
Yes.
left herald
scarlet pond
No…
round parcel
Yes, you still need to differentiate.
round parcel
Use the power rule on each term.
scarlet pond
Bro forgot the mission
round parcel
OK, so let's try (8w^{\frac32}).
grand pondBOT
Chai T. Rex
round parcel
The first thing to do is to multiply by the exponent.
What do you get when you multiply that by 3/2?
left herald
8*3/2?
round parcel
Yes.
left herald
12
left herald
1/2
round parcel
OK, so you have (12w^{\frac12}).
grand pondBOT
Chai T. Rex
left herald
Yes
round parcel
What about (-4w^{\frac12})?
grand pondBOT
Chai T. Rex
left herald
-2w^-1/2 ?
grand pondBOT
Chai T. Rex
left herald
-3w^-3/2 ?
round parcel
Right.
left herald
There we gooooo
round parcel
Now we have (12w^{\frac12} - 2w^{-\frac12} - 3w^{-\frac32}).
grand pondBOT
Chai T. Rex
round parcel
You're welcome.
left herald
I just have one more
slightly similar slightly different
well actually just a quick question
does the 3 root x on the top just simplify to 3x^1/2 ?
round parcel
Yes.
left herald
Okay thank you very much
round parcel
No problem.
round parcel
Do you remember PEMDAS or BODMAS?
round parcel
OK, let's say that x was 4.
If you wanted to do the calculations to find out what number it reduced to, what would the last operation you did be?
From the question, I mean.
left herald
the subtractioin?
grand pondBOT
Chai T. Rex
grand pondBOT
Chai T. Rex
grand pondBOT
Chai T. Rex
round parcel
Nope.
left herald
how
round parcel
The last operation is very important.
left herald
wdym
round parcel
The last thing we do there is the multiplication.
left herald
Yes
round parcel
So, this is a case where we do the product rule instead of the chain rule.
Chain rule is like when you square something as the last step or you take the square root or you do f(something).
The last step is some kind of function.
With one argument, I mean.
left herald
so are my f' and g' x^1/2 and 1?
round parcel
(\sqrt{x} = x^{\frac12} \to \frac12 x^{-\frac12}) for f'.
grand pondBOT
Chai T. Rex
left herald
ohhhh
round parcel
(x - 8 \to 1 - 0 = 1) for g'.
grand pondBOT
Chai T. Rex
left herald
i was just simplifying not taking the derivative
my bad
how deos this look
just simlpify after this?
left herald
ah yeah i got it mixed with quotient
just switched the last two signs, is that final?
round parcel
You can simplify it a bit.
left herald
By adding the first two terms?
round parcel
Yes, and you can also change back to roots.
left herald
Yes got it, thank you
round parcel
You're welcome.
round parcel
OK, so what's the last thing you do with PEMDAS?
left herald
addition
round parcel
That's the first thing you do.
left herald
Okay yeah i got the first t
How is addition not the last here
if theres no subtraction
round parcel
Because that's inside the square root.
So, you need to do t + 9 before the square root can be done.
Let's say that t = 7.
left herald
Because that's inside the square root.
got it
round parcel
And then the square root is inside the logarithm.
So, the square root isn't the last thing.
left herald
is it the log
round parcel
Well, if you get the log, you still need to multiply it by the (t^{\frac32}) you said to do first.
grand pondBOT
Chai T. Rex
round parcel
So, we get the power.
We get the logarithm.
And we multiply them.
Does that make sense?
left herald
So, we get the power.
as in t^3/2?
round parcel
Yes.
left herald
So in the future, if its two seperate terms multiplying id just use product rule
dumb question bc thats what aproduct is '
round parcel
If that's the last operation to calculate it if you know the variable values.
left herald
but like this question and the one above were the same with two terms
round parcel
Let's make it a bit easier to see.
(t^{\frac32}\log_2\sqrt{t + 9} \to \text{pow}(t)\log(\text{sqrt}(t + 9)))
grand pondBOT
Chai T. Rex
round parcel
We're doing it like f(something).
left herald
im confused
round parcel
Like you've seen f(x) and maybe f(2) or f(10) or whatever, right?
left herald
yes
round parcel
OK, so that's a way of writing that we're going to apply a function to a number inside the parentheses.
Like f(9) means to apply the function f to the number 9.
Does that make sense?
round parcel
Right.
left herald
ok
round parcel
First we take the square root of t + 9.
Then we take the logarithm of that.
Then we do the 3/2 power of t.
Then we have two numbers multiplied together.
left herald
then product rule?
round parcel
Right.
But the parentheses make it easy to see what's the last step.
You have to apply the functions to get rid of the parentheses.
So, sqrt can't be the last step because it's inside parentheses.
left herald
Yeah
round parcel
t + 9 can't be the last step for the same reason.
left herald
so root(t) and root9
left herald
how
round parcel
Remember that Parentheses are the first thing to do in PEMDAS.
left herald
what is root(t+9)
round parcel
Well, we're thinking about it as if we knew t.
What's the last step we'd do if we knew t.
Like if t = 7.
Then root(t + 9) = root(7 + 9) = root(16) = 4.
left herald
yeah
round parcel
So, log(sqrt(t + 9)) = log(sqrt(16)) = log(4) = 2.
That's how you have to do it at first.
left herald
howd u get the sqrt16
round parcel
t + 9 = 16 when t is 7.
left herald
but how do we konw t
round parcel
We're just picking a random number.
left herald
is that what you do to solve it?
left herald
So, log(sqrt(t + 9)) = log(sqrt(16)) = log(4) = 2.
I understand this
round parcel
That's what you do to see the order of the steps you follow with PEMDAS.
You can't do sqrt(t + 9) with t left as a variable.
You can't say that's like 27.
So you pick a number for t.
left herald
Oh
round parcel
That lets you go through the PEMDAS steps.
left herald
it wont affect the answer later on?
round parcel
Right, it won't.
left herald
oh i was unaware
round parcel
When the functions get a bit confusing, figuring out the last step in PEMDAS is important in figuring out which rule to use.
round parcel
Right, which is 2.
round parcel
So, we have [t^{\frac32} \cdot \log_2(\sqrt{t + 9})] [7^{\frac32} \cdot \log_2(\sqrt{7 + 9})] [18.52 \cdot \log_2(\sqrt{16})] [18.52 \cdot \log_2(4)] [18.52 \cdot 2]
grand pondBOT
Chai T. Rex
round parcel
Nope, that just tells us the last step you take with PEMDAS.
left herald
Im confused with that
round parcel
It's multiplication, so we use the product rule.
left herald
okay
round parcel
That's the only information we get from it. What rule to use.
left herald
**Chai T. Rex**
From the top, right?
round parcel
No, at the last step where it's 18.52 * 2.
left herald
i use those for product rule?
round parcel
Nope, then you forget all the PEMDAS work.
The PEMDAS work tells us which rule to use.
Other than that, it's useless.
So, we did the PEMDAS work.
We know to use the product rule.
left herald
I think im getting confused with the pemdas
i understand what it does but it gets me mixed up with finding the derivative
round parcel
Right, it's sort of like you take out a separate piece of paper to do the PEMDAS, then you throw it away.
left herald
can you do it without it though?
round parcel
Well, you can learn to see what's done last in PEMDAS without having to do all that work.
It's a bit like solving equations.
Let's say you have (\frac{8x}{5} = 32).
grand pondBOT
Chai T. Rex
round parcel
Well, the last thing you do on the left side is divide by 5.
So, you can multiply both sides of the equation by 5.
(8x = 160)
grand pondBOT
Chai T. Rex
left herald
yes
left herald
i gues since this is relatively new to me i dont pick it up as well as i would with that
round parcel
Have you done that figuring out the last step with equations before?
left herald
wdym
round parcel
Well, you saw that I did a few steps to solve the equation, right?
round parcel
And you saw that I figured out what the last step in calculating the left side was, right?
Like the last thing was initially dividing by 5.
Then the last thing was multiplying by 8.
Let's try this.
[\frac{16x}3 + 6 = 0]
grand pondBOT
Chai T. Rex
left herald
do u mean divide by 8
round parcel
How would you solve this?
round parcel
Right.
left herald
x= blank
round parcel
That's the thing I'm talking about with the product rule and the chain rule.
Like when you figured out to subtract six from both sides, that's because you would add six last.
Like let's say x was 10.
round parcel
Right, you add last.
left herald
so how does this refer back to the original
snow dawn
If anything is equals to 0, the most basic value of x would be 0. Amirite?
round parcel
OK, so you have (t^{\frac32} \log_2(\sqrt{t + 9})).
grand pondBOT
Chai T. Rex
round parcel
The last thing you do is to multiply (t^{\frac32}) by (\log_2(\sqrt{t + 9})).
grand pondBOT
Chai T. Rex
round parcel
Does that make sense?
left herald
yes
round parcel
So, that means you use the product rule.
round parcel
Yes, and (f) is (t^{\frac32}).
grand pondBOT
Chai T. Rex
left herald
yes
round parcel
No.
left herald
uh oh
round parcel
First, do f' because that's the easy part.
round parcel
Oh, OK.
OK, good.
Now, (g = \log_2(\sqrt{t + 9})). What's the last step in calculating that?
grand pondBOT
Chai T. Rex
left herald
addition
round parcel
Nope.
left herald
parenthesis
round parcel
That's what you do first.
Like let's say you were solving for t.
(\log_2(\sqrt{t + 9}) = 2)
grand pondBOT
Chai T. Rex
round parcel
What would be your first step?
left herald
the log?
round parcel
Right, you'd exponentiate both sides.
[2^{\log_2(\sqrt{t + 9})} = 2^2]
[\sqrt{t + 9} = 4]
grand pondBOT
Chai T. Rex
round parcel
You can also tell because it's the function surrounding everything else.
left herald
**Chai T. Rex**
wait why =2?
round parcel
As an example.
left herald
ok
round parcel
Figuring out which differentiation rule to use is a lot like figuring out the next step in solving an equation.
So, we have log_2.
That's a function.
It takes one number into it.
So we use the chain rule.
Notice also that everything is inside the log here.
(\log_2(\sqrt{t + 9}))
grand pondBOT
Chai T. Rex
round parcel
When you see that, you use the chain rule.
It's a function with everything else inside of it.
left herald
Okay
so can i ask this
for this, is rootx not a function since the root doesnt extend through the parenthesis?
round parcel
It is a function, but it's easier to solve it using the power rule.
round parcel
Wait, what are you asking? I think I misunderstood.
round parcel
Oh, OK.
You don't use the chain rule for the (\sqrt{x}(x - 8)) because of what you said. Only part of everything is inside the square root.
grand pondBOT
Chai T. Rex
left herald
Got it
round parcel
Let's say you were solving ((x^2 + 2x + 3)^2).
grand pondBOT
Chai T. Rex
round parcel
Everything is inside the squared parentheses.
So you can use the chain rule with that.
left herald
Yea
round parcel
OK, let's go back to (\log_2(\sqrt{t + 9})).
grand pondBOT
Chai T. Rex
round parcel
How would the chain rule go for this?
left herald
1 / (2(t+9)ln2)
round parcel
OK, so the chain rule has two parts.
They're multiplied together.
The first one leaves the stuff inside the log alone and differentiates just the log part.
left herald
Yes
round parcel
Like (\frac1{\sqrt{t + 9} \ln(2)}).
grand pondBOT
Chai T. Rex
round parcel
The other part is just the derivative of the stuff inside.
What's the derivative of (\sqrt{t + 9})?
grand pondBOT
Chai T. Rex
left herald
1/2root(t+9)
round parcel
OK, so we have (\frac1{\sqrt{t + 9}\ln(2)} \cdot \frac1{2\sqrt{t + 9}}).
grand pondBOT
Chai T. Rex
round parcel
So you were right.
round parcel
OK, so now we can go back.
round parcel
[f = t^{\frac32}] [f' = \frac32 t^{\frac12}] [g = \log_2(\sqrt{t + 9})] [g' = \frac1{2 \ln(2) (t + 9)}]
grand pondBOT
Chai T. Rex
round parcel
Oh, OK.
round parcel
Yep, looks good.
left herald
Cool, thanks a lot for the help
round parcel
You're welcome.
midnight plankBOT
round parcel
You too.
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chilly cobalt
Find all nonempty finite sets of positive integers ${a_1,a_2,...,a_n}$ such that $\ \prod_{i=1}^{n}a_i$ divides $\prod_{j=1}^{n}(x + a_j)$ for every positive integer $x$
midnight plankBOT
Find all nonempty finite sets of positive integers $\{a_1,a_2,...,a_n\}$ such th...
grand pondBOT
Copter
chilly cobalt
i was given a hint to use discrete difference for this problem... still dont know where to start ;-;
oh wait nvm
ts was easy
.close
midnight plankBOT
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inland patio
Suppose I have some function $f:X\to Y$ and a closed set $E\subset X$. Is it true that if $x\notin E$, then $f(x)\notin \overline{f(E)}$?\
Attempt: Suppose $f(x)\in\overline{f(E)}$. Then it is either in $f(E)$, which it can't be because $x\notin E$, or it is an accumulation point of $f(E)$. If it is an accumulation point, then $f(E)\cap (U\setminus{f(x)})\neq\varnothing$. How do I continue from here?
midnight plankBOT
Suppose I have some function $f:X\to Y$ and a closed set $E\subset X$. Is it tru...
grand pondBOT
psie
visual tiger
I would start by considering a constant function xd
Your logic fails at "Then it is either in f(E), which it can't be because x is not in E"
As f^-1(f(E)) ≠ E
Maybe an injective continuous function would work
Or just injective function, idk haven't considered it yet
inland patio
Maybe an injective continuous function would work
Yes, f is continuous and injective. 😄 I didn't mention that, but it is.
inland patio
As f^-1(f(E)) ≠ E
Then this is true I think, or?
visual tiger
Then f^-1(f(E)) = E as f is injective
inland patio
Ok, good. The case where f(x) is an accumulation point remains.
visual tiger
Ok, good. The case where f(x) is an accumulation point remains.
The statement you wrote is true for all U open neighborhood of f(x)
inland patio
The statement you wrote is true for all U open neighborhood of f(x)
Yes.
inland patio
The statement you wrote is true for all U open neighborhood of f(x)
Doesn't that show f(x) is in the closure of f(X)? What you said is another characterization of a point being in the closure.
visual tiger
Actually just wondering
inland patio
Ok. I feel like in the case where f(x) is an accumulation point, one perhaps has to use that E is closed, not sure.
midnight plankBOT
@inland patio Has your question been resolved?
midnight plankBOT
@inland patio Has your question been resolved?
inland patio
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molten bay
2|x|+3|y|<=6
midnight plankBOT
2|x|+3|y|<=6
lyric charm
make a graph
molten bay
But I was searching for a solution which I lost somewhere on math exchange website i guess
lyric charm
start with first quadrant then mirror across both axes
molten bay
It was like direction of a line
midnight plankBOT
@molten bay Has your question been resolved?
midnight plankBOT
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fair frigate
is ln just log with base of e
midnight plankBOT
is ln just log with base of e
somber plover
is ln just log with base of e
yes
hushed mauve
correct
midnight plankBOT
hushed mauve
no stupid questions
vernal token
Given that sin(x) + cos(x) = 1/5, calculate sin(x), cos(x), tan(x), and cot(x). helppp
hushed mauve
think this channel is still not yet unbound
fair frigate
maybe
vernal token
Wait, is this server still running?
steel crest
what makes you think it stopped
asian math nerd typed ".close"
hushed mauve
asian math nerd typed ".close"
i mean at best it stops the channel, not the server
steel crest
oh, i didn't even register the word "server"
midnight plankBOT
Available help channel!
midnight plankBOT
turbid cairn
.close
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final burrow
hi
delicate cradle
lyric charm
hi
hello, welcome to the server!
do you have a math question to ask, or did you just want to chat?
if you just want to chat, #discussion and #chill are our channels for that.
midnight plankBOT
@final burrow Has your question been resolved?
rich mirage
fallen vault
Does the integral of 1/x between -1 and 3 diverge or is it equal to ln(3)?
I thought that improper integrals of this type have value only if the integrals that make it (in this case, from -1 to 0- and from 0+ to 3) DO NOT diverge. But in this case they both diverge, but https://www.integral-calculator.com claimes it's equal to ln(3), although wolfram alpha does say that the integral diverges.
Is my understanding correct and I just found the limits of that website? Should I only trust wolfram alpha from now on?
midnight plankBOT
Does the integral of 1/x between -1 and 3 diverge or is it equal to ln(3)?
I tho...
carmine sigil
It diverges, but it has a Cauchy principle value which is probably equal to what you wrote.
fallen vault
Cauchy principle value doesn't mean that you can in this case divide the integral into the two I mentioned in order to find the value?
.close
midnight plankBOT
carmine sigil
Cauchy principle value doesn't mean that you can in this case divide the integra...
Cauchy principle value means you can find a limit because the two sides of the infinity tend to cancel out.
this is made clear by the fact that you used 0+ and 0-
these are limits that approach 0, but do not contain 0.
Hope this helps @fallen vault 😄
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cerulean halo
Does the integral of 1/x between -1 and 3 diverge or is it equal to ln(3)?
I tho...
integral-calculator says it diverges
midnight plankBOT
integral-calculator says it diverges
steep condor
1.1.4 , b)
midnight plankBOT
1.1.4 , b)
runic hamlet
what have you tried
steep condor
lol
literally
tried adding
i considered two coordinates
on the circle
theta and phi
yeah thats what i tried
lol
runic hamlet
what did you get
steep condor
a vector with coordinates $$ (\cos{x} + \cos{y},\sin{x} + \sin{y} ) $$
L
anyways
yeah
something like that
idk what to do after that
runic hamlet
to be on the unit circle what equation would the new point have to satisfy
grand pondBOT
grey
steep condor
to be on the unit circle what equation would the new point have to satisfy
oh yes
it cannot be on the circle because we would not get the sum of their squares equal to 1
runic hamlet
why not
steep condor
we can actually nvm i got the sum of their squares as
4cos^2((x-y)/2)
when we equate that to 1
we can get some solutions for x and y
so answer to the problem
would be
YES
am i right ?
@runic hamlet
runic hamlet
yes
a different way to think about it is that you can imagine rotating the setup
so if two of those vectors exist, we can rotate one of them to be (1,0)
and then we can more explicitly solve for the other one
steep condor
oh
yes
it would make it much easier
thanks
its my first time reading a math book
lol
.close
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lament knoll
any free family of cardinal $n$ in a vector space of dimension $n$ is a basis.
We use the incomplete basis theorem to eventually complete this free family with vectors. However, it has cardinal n, so if we add vectors to it, it won’t be a basis, which is excluded.
how can we directly deduce from this that this family is a basis of $E$?
grand pondBOT
tm
midnight plankBOT
any free family of cardinal $n$ in a vector space of dimension $n$ is a basis.
...
lament knoll
like ok we can’t complete this family but why it’s automatically a basis
runic hamlet
we are assuming n is finite?
lament knoll
yes ofc
runic hamlet
well you were the one who wrote cardinal
just an unusual word choice
well anyway we need to show linearly independent and generating
linearly independent is given
if there was a vector not in the span, we could add it
but we cant
so its generating
midnight plankBOT
grim vector
any free family of cardinal $n$ in a vector space of dimension $n$ is a basis.
...
ça va le t ?
lament knoll
trankilou ça reprend l’algèbre linéaire depuis le début mdr
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quaint shoal
When graphing this, (top left)
midnight plankBOT
When graphing this, (top left)
quaint shoal
Ik I need to find the slant asymptote
As the tops degree Is larger than the bottoms, but which 2 terms would I divide?
grim vector
Ik I need to find the slant asymptote
what does it give ?
quaint shoal
what does it give ?
wdym?
grim vector
the slant asymptote
quaint shoal
the slant asymptote
oh it helps you graph the equation
grim vector
yeah so what is it ?
quaint shoal
some sort of linear line
quaint shoal
yeah so what is it ?
This is the graph for the
grim vector
ok but if you need it, do it
quaint shoal
ok but if you need it, do it
that’s the plan but idk how 😭
grim vector
ah ok
tell me this ima explain you
ok so to find slant asymptote you need to calculate two limits
- f(x)/x in +inf = some real m
- f(x) - mx = some real p
quaint shoal
I haven’t done limits
grim vector
your slant asymptote is y = mx+p then
quaint shoal
the teacher said all u do is divide the equation out
grim vector
hmm
quaint shoal
I got this
Here
which seems pretty easy, but then going back to my original equation, do I just divide the original equation or the simplified ones?
quaint shoal
so in your case its saying that there is no slant asymptote ?
I think it means no horizontal asymptote
grim vector
oh ok
quaint shoal
i actually find wierd to do slant asymptote without limits
these aren’t even in the course book and he just added it to “prepare us”
😭
same question
grim vector
also becareful when posting, especially to your name
quaint shoal
but this is wrong from what is written
why? The degree is 3 on top and 2 on the bottom
3 is larger so it’s slant?
grim vector
why? The degree is 3 on top and 2 on the bottom
i say that the propreties is either wrong or typo
grim vector
Here
here
n > m isn't always slant
Slant asymptotes are caused by the numerator having a degree that is 1 greater than that of the denominator; they indicate where the graph will be when it's off to the sides.
so you should rather write m+1 = n
grim vector
if i tell you (x^4 + 1)/(x^2 -4)
counter example :
quaint shoal
Slant asymptotes are caused by the numerator having a degree that is 1 greater t...
hmm
here’s what I did to find it
grim vector
polynomial division?
quaint shoal
polynomial division?
yea
Ik how to do that
But I just don’t know where the x^2 -2x +5 /x+4 comes about
grim vector
this is working yes
quaint shoal
When graphing this, (top left)
seems like I used the simplified versions(middle one on top)
But just excluded (x) and (-4x)
quaint shoal
But just excluded (x) and (-4x)
I’m just not sure why u wouldn’t include these 2 numbers when dividing
midnight plankBOT
@quaint shoal Has your question been resolved?
quaint shoal
<@&286206848099549185>
fallow scarab
I’m just not sure why u wouldn’t include these 2 numbers when dividing
x / (-4x) = -1/4 for all nonzero x. so you can cancel and do the poly division on the rest then multiply by -1/4 at the end
quaint shoal
x / (-4x) = -1/4 for all nonzero x. so you can cancel and do the poly division o...
Oh I see
ok I think I see where it comes from
thanksss!
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muted patio
trynna find some interesting vids to learn awesome mathematical concepts. also, if you could, please tell me some free webs like khan academy to learn maths 🙏 so I want to familiarise myself with math olympiad exercises, physics, and advanced math, but I don’t see how to advance yk
midnight plankBOT
trynna find some interesting vids to learn awesome mathematical concepts. also, ...
raven zenith
civic lynx
3blue1brown's whole shtick is creating and explaining geometric interpretations of concepts. he's done the fourier transform and convolution, as an example, and quite a bit of calc I concepts too
muted patio
and like for learning math stuff, khan academy or other webs?
civic lynx
#book-recommendations has some good website recommendations, contrary to its name. check pins
muted patio
<#716264872018706443> has some good website recommendations, contrary to its nam...
oh thanks, thought it was only books
civic lynx
lol thats fair, there's a bunch of stuff :) and there are some book recommendations for undergrads/beginners
muted patio
lol thats fair, there's a bunch of stuff :) and there are some book recommendati...
thanks, imma check that out!
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raven zenith
<@&268886789983436800>
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rustic garden
$$\lim_{x \to \infty} \left( \sum_{k=0}^{x-1}(-1)^k \binom{z}{k} (x-k)^z \right)= \Gamma(z+1)$$
midnight plankBOT
$$\lim_{x \to \infty} \left( \sum_{k=0}^{x-1}(-1)^k \binom{z}{k} (x-k)^z \right)...
grand pondBOT
Cozmogrgdfschkipkhrshtensi
rustic garden
I know that this limit agrees with the gamma function where it converges, but I would like to show that it converges for all complex z where Re(z) > -1
I tried converting the sum to an integral, splitting it in various ways, and it always seem to be a little bit out of reach of any methods I can toss at it, anyone have any ideas?
basically, as long as I can show it converges, I think I can show it is analytic, and then by Carlson's theorem it will match everywhere
<@&286206848099549185>
carmine sigil
How are you computing the binomial coefficient on complex z?
(are you sure you wrote down the sum correctly?)
rustic garden
I'm doing the falling factorial divided by k!
treating it as a polynomial of degree k
I did try turning it into a bunch of gamma functions but then I couldn't figure out how to turn the sum into an integral, so that felt like a dead end
rustic garden
(are you sure you wrote down the sum correctly?)
ive "reasonably" verified the sum computationally
i set x to be like 10,000 and it seems to converge fairly well to where I think it should converge, and it pretty clearly looks like it's diverging elsewhere
midnight plankBOT
@rustic garden Has your question been resolved?
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@rustic garden Has your question been resolved?
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@rustic garden Has your question been resolved?
fallow scarab
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@rustic garden Has your question been resolved?
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viral dagger
prove that there does not exist an integer solution to $4x^3-7y^3=2003$
midnight plankBOT
prove that there does not exist an integer solution to $4x^3-7y^3=2003$
grand pondBOT
skissue.in.a.teacup
viral dagger
mod 8 didnt seem promising
mod 9 yielded that x≡2 mod 3 and y≡0 mod 3
subbing back yielded this
err pretend its k instead of x
midnight plankBOT
chilly cobalt
rare copter w
fathom onyx
4(-1,0,1)=1 mod 7
Phrasing is slightly off
viral dagger
huh
viral dagger
rare copter w
"rare" they said
fathom onyx
Testing for $x \in {0,1,2,3,4,5,6}$, then $x^3 \in {0, 1, 6} \mod 7$
viral dagger
yes x^3 is either -1,0,1
grand pondBOT
Waes (Wires)
fathom onyx
which is a contradiction, so there are no solutions
fathom onyx
4(-1,0,1)=1 mod 7
I just mean that this line is a little vague, especially for a "prove that" question, is all
You've otherwise got it 🫡
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Available help channel!
chilly cobalt
let $n$ be a positive integer. Show that there exists a polynomial $P(x)$ with integer coefficients that satisfies:
$\ |P(x)| = (k-1)!(2^n-k)!$ for each $k \in {1,2,..,2^k} \$ Degree of $P$ is at most $2^n - n - 1$
midnight plankBOT
let $n$ be a positive integer. Show that there exists a polynomial $P(x)$ with i...
grand pondBOT
Copter
chilly cobalt
this is all i have sadge
lavish gyro
Guys is anyone here available for tutoring?
gaunt nimbus
!occupied
midnight plankBOT
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).
gaunt nimbus
Guys is anyone here available for tutoring?
If u have a problem, plz post in another channel
lavish gyro
If u have a problem, plz post in another channel
Okiee
chilly cobalt
**Copter**
- i dont think "P with integer coefficients" is necessary cause we can prove it must be
<@&286206848099549185> mahoraga helpp
midnight plankBOT
@chilly cobalt Has your question been resolved?
midnight plankBOT
@chilly cobalt Has your question been resolved?
chilly cobalt
sadge
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visual tinsel
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visual tinsel
I tried solving this by using the circle equation
(x-k)^2 + (y-h)^2 = r^2
(x-6)^2 + (y-h)^2 = 81
then i plugged in a point from the x-axis
but i dont get anything that resembles the answer choices
blissful pier
Instead, I'd suggest you observe that the distance from C to (3,0) and (15,0) will both be 9 units as those are both radii. Using this fact we have two unknowns (m,n) and two equations
visual tinsel
so use the distance formula twice?
using 9 as the distance, and the zeros as the points?
whoa
makes sense
thanks mama llama
emojis of joy and happiness
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blissful pier
emojis of joy and happiness
emojis of sugar and spice and everything nice
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✅
visual tinsel
im back because i got stuck again
when i use 2 distance formulas, one of them equals to 9 = Square Root(81 + n^2) and the other one equals 9 = Square root(9 + n^2)
you can't these 2 equal to each other, since it just comes out to no solutions
Been 8 minutes, 7 more to go
Until i can ping
Sigh
cerulean solar
?
visual tinsel
nice
cerulean solar
I’m confused about what you did for your work
visual tinsel
there are 2 distance equations i can use, just as llama said
distance formula is d = Square root ((x2 - x1)^2 + (y2 - y1)^2)
i set distance to 9, and used the point (15,0) and (3,0)
steel crest
when i use 2 distance formulas, one of them equals to 9 = Square Root(81 + n^2) ...
can we have some context
visual tinsel
i got 2 equations with 2 unknowns, n and m
visual tinsel
when i use 2 distance formulas, one of them equals to 9 = Square Root(81 + n^2) ...
The equations i got were these .
steel crest
you replaced m with 6 in what?
visual tinsel
the distance formulas i used
cerulean solar
?
visual tinsel
I will send picture of my work
that should maybe make it more clear
one sec
i have a flip phone, so i have to use my webcam on my computer
does this make sense?
cerulean solar
,rotate
grand pondBOT
cerulean solar
Where did the 6 come from
visual tinsel
its the middle of 3,0 and 15,0
since the middle of that has to be the mid point
which is where the center would lie
cerulean solar
I see
N measures the y
And there are two points that are equal distant of 9 units away from the two zero points
visual tinsel
LlamaMama (the guy previously) helped me realize this part
I have 2 minutes left to solve this problem ;-;
or else i exceed time, and that means i study more than i have to
but i wanna solve this really bad
not relevant to the problem
sorry
cerulean solar
You’re on the right track, you just gotta find the Ns now
visual tinsel
yeah, but when i set these 2 equations equal to each other
it doesnt work
since Square root (81 + n^2) can't be equal to Square root (9 + n^2)
visual tinsel
it isn't?
cerulean solar
That’s why you’re equations are wrong
cerulean solar
Yes
visual tinsel
...
I love making simple mistakes
thanks for the help sir
have a good day
.close
woeful turret
midnight plankBOT
blissful pier
yeah this is one of those questions where one trick makes it easy
notice to what variable you are differentiating with respect to :)
blissful pier
assuming it isn't a typo, and since it's not specified otherwise
theta is independent of x
so it's a constant
so this whole nasty expression for y... is also a constant wrt x
blissful pier
twilit jetty
sorry but the answer is 1
simplify y to get 1
then d/dx of that is a constant is 0
blissful pier
right... the answer is 0, no?
twilit jetty
you can simplify y with algebra
blissful pier
oh wait I see what ur saying alright
yeah but still noticing no x in y(theta)
just gives 0 immediately
woeful turret
you can simplify y with algebra
oh yes..the expression comes out nicely
sorry guys the questions are usually more tricky than this
so i didnt notice
ty
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twilit jetty
half of this question is just being extremely dense to read
woeful turret
yeah 😔
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neat silo
$\floor{2x} + \ceil{5x} = 2025$
midnight plankBOT
$\floor{2x} + \ceil{5x} = 2025$
grand pondBOT
isuckatmath
neat silo
I did casework for x is an integer and x is not an integer
For x an integer
$2x + 5x = 2025$ no solutions
grand pondBOT
isuckatmath
neat silo
For $x$ non integrr
$\floor{2x} + \ceil{5x} = \floor{2x} + \floor{5x} + 1$
grand pondBOT
isuckatmath
neat silo
So its just $\floor{2x} + \floor{5x} = 2024$
grand pondBOT
isuckatmath
neat silo
How do i solve this?
feral sedge
$\floor{2x} + \ceil{5x} = 2025$
you consider the behavior of floor(2x) + ceil(5x) around x = 2025/7
this is fine because floor(2x) + ceil(5x) is an increasing function
floor(2x) and ceil(5x) are integers so you consider which are possible integers floor(2x) and ceil(5x) so their sum is 2025, and what range of x gives those
neat silo
you consider the behavior of floor(2x) + ceil(5x) around x = 2025/7
this is fin...
is x = 2025/7 the only sol?
feral sedge
definitely not
feral sedge
there is no need to reframe the problem
neat silo
into like 7 parts and do casework
feral sedge
floor(2x) and ceil(5x) are integers so you consider which are possible integers ...
you do this
feral sedge
in particular you note that floor(2x) and ceil(5x) when x = 2025/7 are 578 and 1447, which do sum to 2025
neat silo
yes
feral sedge
so a good thing to try is to find for which x you have floor(2x) = 578 and ceil(5x) = 1447
neat silo
oh
floor(2x) = 578 bound this
ceil(5x) = 1447, bound this
then find the intersection
?
feral sedge
yes essentially
neat silo
floor(2x) = 578 bound this
this becomes 578 <= 2x < 579
feral sedge
yes essentially
but you may need to consider other pairs of integers i.e. 577, 1448 or 579, 1446
neat silo
feral sedge
- not an interval
- even if presented properly, this is wrong
- you are helping nobody by simply posting an answer
feral sedge
floor(2x) = 578 bound this
this becomes 578 <= 2x < 579
seems fine, you can continue making a bunch of bounds and seeing what happens
the point of this problem is you have to find the most restrictive bounds that give you a correct sum
neat silo
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neat silo
yes
midnight plankBOT
Available help channel!
sleek marsh
Can someone help
midnight plankBOT
Can someone help
gaunt plume
What have you tried?
sleek marsh
I got 15km/h faster=1/3h faster
gaunt plume
Probably easier to keep it in minutes rather than hours but it doesn't really matter
What you have to do now is find some way of relating the time he takes to the distance he travels
gaunt plume
Probably easier to keep it in minutes rather than hours but it doesn't really ma...
Actually hours might be easier but like I said it doesn't make much difference
gaunt plume
It's how I'd solve it
imo it's the simplest way but there are always multipe ways of doing it
But basically, let D be the distance from his house to his office, and let T be the time it takes him at 60 km/h
Then you should be able to write down the info in the question as two simultaneous equations in terms of D and T
(which you can then solve)
finite wharf
@gaunt plume how well are you with alg 101?
gaunt plume
Is that a specific american thing? bc if so then idk what that is
I am a postgrad though so I know more than enough algebra to solve a question like this
midnight plankBOT
@sleek marsh Has your question been resolved?
finite wharf
Is that a specific american thing? bc if so then idk what that is
no i meant, cause i could use some help with my own question lol
but its resolved now
viral dagger
no i meant, cause i could use some help with my own question lol
fyi, advertizing your help channels is againts the servers rules