Like 1/2 and 2 don’t make y=2.5
#help-19
frosty stone
brittle beacon
,calc 2 + (1/2)
clever fjordBOT
Result:
2.5
,calc 0.5 + (1/0.5)
clever fjordBOT
Result:
2.5
Put those into the y equation
You want where the parametric curve meets y= 2.5, so set the y equation equal to that 2.5, so you find those values 
frosty stone
Okay 1 + 2 / 2 = 3
brittle beacon
Remember the $y$ equation is $y = t + \frac1t$, so if you have $t = 2$, or $t = \frac12$, you should get 2.5
clever fjordBOT
@brittle beacon
brittle beacon
How do you mean combined?
brittle beacon
As in making it one fraction?
frosty stone
Yes
brittle beacon
If so, that's $\frac{t^2 + 1}{t}$
clever fjordBOT
@brittle beacon
brittle beacon
Peak 
happens 
frosty stone
Alright thank you literally been stuck on this for like 30 minutes so confused lmao
Imma try it on my own now
.close
frosty stone
I’m confused why the bounds change here when you plug in
brittle beacon
The same 32 right?
brittle beacon
well remember that it's a similar "substitution" thing - to be fair, not entirelt sure why they care about the x limits, because you don't really need them, probably they put them for completeness
Of course the "original" integral is of the difference between "the top curve" and "bottom curve", but with everything in terms of x
frosty stone
Wdym so you say you could’ve got the answer without switching bounds again
brittle beacon
But then of course you want to convert that to parametric form
frosty stone
What do you mean don’t you just evaluate the integral
brittle beacon
Wdym so you say you could’ve got the answer without switching bounds again
Well I mean - technically you need to switch the bounds for the integral $\int (2.5 - y) \dd y$, but the most important is that you only care about working out $\int_{\frac12}^2 (2.5 - y)\dv{x}{t} \dd t$
clever fjordBOT
@brittle beacon
brittle beacon
Like would this be good to go
Yep - that's like what you want for the actual area 
frosty stone
Okay
I don’t care about the notation that much I was just confused again lol
I was like wtf
Thank you
.close
fair oriole
I dont know how to solve this, can someone explain to me what i have to do ?
fair oriole
Can you draw A"B"C"?
im kind of confused on how to draw reflection over x = -1
how do i draw that ?
drowsy root
It means that if you had a mirror on x=-1
Where would those points appear
For example the point (-3,0) reflected along (-1,0) would be (1,0)
drowsy root
Not quite
drowsy root
The y value isnt correct
drowsy root
Exactly
fair oriole
ah
so it changes
which ever it has
like if it was y =
it wouldve changed the Y value ?
drowsy root
Now what about the other points?
Yes?
If a point is on the line, it wouldnt change
Perhaps its easier to see it like this
fair oriole
fair oriole
so it would become 0,-3 ?
drowsy root
That would be a reflection along x=1/2
fair oriole
1,-3 ?
drowsy root
That would be a reflection along x=1
It doesnt change it
Look at the closest point on the line to your point
For B that would be?
fair oriole
like closest to where it originally was ?
drowsy root
The closest point on the reflection line x=-1 and point B
fair oriole
-2, -3 ?
drowsy root
Is -2,-3 on the reflection line x=-1?
Perhaps its easiest of you draw it in the diagram
Draw the point B (1,-3)
Draw the reflection line x=-1
fair oriole
And now what about A?
1, -3
fair oriole
fair oriole
fair oriole
this is the reflected triangle
fair oriole
drowsy root
But it would switch left-right
And the top left corner just went down
When it shouldve actually been reflected and gone to the right
drowsy root
But its a reflection along x=0
Do you see why this is the case?
fair oriole
Do you see why this is the case?
Didnt i find the right values earlier
?
r were they wrong
drowsy root
A"=(1,3), B"=(-3,-3), C"=(1,-3)
Draw these points
Do you see why this is the reflection along x=-1?
fair oriole
they are all reflected above the X line ?
drowsy root
Draw the points
fair oriole
fair oriole
drowsy root
One moment
(1,3),(1,-3),(-3,-3)
These points
Draw these points and the triangle
Then draw the original triangle
And then also the line of reflection
fair oriole
drowsy root
Coincidentally, yes
Wait
Im gonna give you
Some points
And you tell me
Where they reflected from
Okay?
fair oriole
ok
drowsy root
(3,-3),(7,-3),(7,3)
fair oriole
those are the reflected points and you want me to tell you the orginal points ?
or this are the og points
these
drowsy root
No, use the og points you have
Snd these are the reflected points
Along which line do they reflect?
X=?
Can you show your drawing?
drowsy root
Your og triangle is wrong
fair oriole
oops
fair oriole
fair oriole
X= 3 ?
drowsy root
Not quite
Look at the midpoint between each sets of points
Look at the midpoint of the top two
The midpoint of the righr angle ones
fair oriole
X = 2 ?
drowsy root
And the midpoint of the other ones
odd edgeBOT
drowsy root
Huh
fair oriole
?
real trellis
ugh
bot outage
you can ping us, its fine
sometimes discord API does a bad and tells the bot that everyone is missing
we are looking into ways to fix this behaviour
odd edgeBOT
Available help channel!
craggy anchor
Where would I start with this?
odd edgeBOT
Where would I start with this?
nocturne belfry
craggy anchor
Not sure if you can open this but https://docs.google.com/document/d/1NFsRdGEajp68sKJDyD_DpuIzJl0TAZHd5YLHXcJRIGA/edit
let me make a copy
true minnow
you can apply e^ln(x) to both sides
craggy anchor
true minnow
then you can apply log rules:
ln(a^b)=b ln(a)
craggy anchor
you can apply e^ln(x) to both sides
Will do
craggy anchor
you can apply e^ln(x) to both sides
Wouldnt the e and ln cancel leaving you where you started?
true minnow
well yes, but we ignore that for now
but yes, that is why we are allowed to do that
if a=b
then e^ln(a)=e^ln(b)
then also ln(a)=ln(b)
so in our case:
a^b=c^d
e^ln(a^b)=e^ln(c^d)
e^bln(a)=e^dln(c)
b ln(a) = d ln(c)
that is the idea
craggy anchor
where would I go from there?
true minnow
when we have it in the form
b ln(a) = d ln(c)
we try to factor out the x
and then we try to simplify the rest
craggy anchor
I see
plain badge
I see
@craggy anchor
I can help you solve
without log
just use the laws of exponents
plain badge
(2/3)^(x-5) = (2/3)^(-3x/2) 9/4 = (2/3)^(-2 )
true minnow
then apply (a^b)^c=a^(bc)
odd edgeBOT
true minnow
.reopen
odd edgeBOT
Available help channel!
true minnow
@plain badge
i tried your method
odd edgeBOT
<@886846638730129509>
i tried your method
fringe bough
haven't looked closely enough at the second one yet
plain badge
it will be -3x
@true minnow
you forgot the '-'
(3/2)^(3x/4)(2) = (2/3)^(-)(3x/4)(2)
true minnow
yeah, but what about the other 2 pictures?
plain badge
idk
true minnow
,w solve (2/3)^(x-5)=(9/4)^(3x/4) for x
clever fjordBOT
true minnow
oh god
i mean, that there are complex solutions kinda makes sense when dealing with exponents
true minnow
well it's late, so i will think about this tomorrow haha
have a good night^^
.close
odd edgeBOT
odd edgeBOT
Available help channel!
spice mauve
What is the 15th root of this
odd edgeBOT
What is the 15th root of this
spice mauve
3,699,376,128,000
drowsy root
use a calculator
frigid canopy
😭 I don’t know how to do that on my phone
Use a more advanced calculator
drowsy root
,w calc (3699376128000)^(1/15)
clever fjordBOT
frigid canopy
like wolfram
spark flare
odd edgeBOT
spark flare
I don’t know how to get started, can someone guide me through the question?
halcyon parrot
assume speed of train to be 'x'
find relative velocity b/w train and each of the cyclists
i think that shud be the way to start
odd edgeBOT
@spark flare Has your question been resolved?
spark flare
assume speed of train to be 'x'
find relative velocity b/w train and each of the...
like (x-8.1) and (x-12.6)
?
odd edgeBOT
odd edgeBOT
Available help channel!
icy coral
Hi guys,is this a solvable question?
odd edgeBOT
Hi guys,is this a solvable question?
icy coral
I tried many different ways still cant simplify and the answer is always long and weird
honest turtle
Is this a test?
honest turtle
Alright. You need to factorize and cancel out common terms.
honest turtle
I think they probably did something wrong with some of the signs.
signal oar
I think they probably did something wrong with some of the signs.
Did they?
honest turtle
At least with calculator and my own calculations, I didn't get the expression factorized.
signal oar
Yeah, seems so
honest turtle
Hi guys,is this a solvable question?
No, I don't think so. Skip over the question.
green vine
if the division sign was a multiplication sign i think thatd work
should be a printing error
icy coral
Alr thanks a lot guys
signal oar
,w (x^2 - 3x)/(16x - 8) * (2x^2 - 1)/(4x^3 - 12x) simplify
icy coral
Well ya prob the printing error,thanks!
signal oar
There is something interesting going in here though
honest turtle
Well ya prob the printing error,thanks!
You mind telling the title of the book?
green vine
should we just mulitply and use long division then
signal oar
should we just mulitply and use long division then
There might be some trick here
signal oar
Notice 2x - 1
Using this
signal oar
2x^2+-1???
Yes, that's usually -1..
dusk orbit
Something is missing
icy coral
You mind telling the title of the book?
Nope but my book kinda rotten and the title is gone
signal oar
But usually written as + (-1)
icy coral
Cuz im malaysian,so i expect the book should be form 2 kssm something like that
dusk orbit
I think that's 2x^2+x-1
signal oar
Hi guys,is this a solvable question?
,w simplify (x^2 - 3x)/(16x - 8) * (4x^3 - 12x)/(2x^2 - 1)
We need x + 1 or x - 1 in the numerator
dusk orbit
Yea guessed so
signal oar
I think that's 2x^2+x-1
Hm, let's try with that
,w simplify (x^2 - 3x)/(16x - 8) * (4x^3 - 12x)/(2x^2 + x - 1)
clever fjordBOT
signal oar
Still no
dusk orbit
Well:(
signal oar
Hi guys,is this a solvable question?
At which chapter is this
Were there exercises about completing the square just before this?
Or polynomial long division..
dusk orbit
Maybe its 4x^2
icy coral
At which chapter is this
For malaysian,it is chapter 2 for factorisation and algebraic fractions
icy coral
Were there exercises about completing the square just before this?
It is a kinda like final exercise for all chapter and i dont think the question will be long
signal oar
Maybe its 4x^2
There must be something going on with (x^2 - 3) * (x - 3) and (2x - 1) * (2x^2 - 1)
edgy bay
How do you prove that a polynomial of degree n with n+1 roots must be the zero polynomial? Someone said that this easily follows from the fundamental theorem of algebra (FTA), but I don't see how this follows from the FTA
low locust
this isnt really about FTA
in general a polynomial of degree n can have at most n roots
which you can prove with induction
edgy bay
right
low locust
if f(x) has a root, then it has a factor x-a. so f(x)=(x-a)g(x). because this is a product, its only zero if one of the factors is zero. so if f(x)=0 then either x=a or g(x)=0. note that g(x) is a polynomial of degree n-1. and then induction
edgy bay
hmm, so why does this show that f(x) must be the zero polynomial? or did you try to show something else?
or could one simply argue that the zero polynomial has infinitely many roots?
low locust
it clearly does
well my argument only works for polynomials of degree at least 1
so if f had degree at least 1, then it would have at most n roots
it has more roots, so it has degree < 1, so it is constant
and then it has to be constant zero
edgy bay
it has more roots, so it has degree < 1, so it is constant
could you explain this a bit more "it has more roots, so it has degree < 1" why is that?
low locust
well if it has degree >=1, then by my argument it can have at most n roots. but it has n+1 roots
edgy bay
hmm, is this reasoning the contrapositive of your argument?
low locust
well I presented it more as a contradiction. but yes
edgy bay
.close
odd edgeBOT
odd edgeBOT
Available help channel!
signal oar
odd edgeBOT
signal oar
It's because we squared, so we need to plug back and check, right?
But how could we check
Ah
Using that roots are always positive
.close
odd edgeBOT
odd edgeBOT
Available help channel!
finite birch
How did the teacher somehow do 1.414 times p?
odd edgeBOT
How did the teacher somehow do 1.414 times p?
finite birch
the 1.414 is understandable
assume Faf is 2kN and multiply it by sin(45)
but how tf did he get 1.414 times P?
honest turtle
1/sin(45) = 1.4142
finite birch
1/sin(45) = 1.4142
wait but why divise by 1
honest turtle
Since you're isolating F_AF.
finite birch
Since you're isolating F_AF.
but to divise by 1 you'd have to assume that P = 1
honest turtle
$$F_{AF} \cdot sin(45)-p=0$$ $$F_{AF} \cdot sin(45)=p$$ $$F_{AF}=\frac{p}{sin(45)}$$
clever fjordBOT
Good
finite birch
oh so you're just assuming the p has a coeffecient of 1
ok
understandable
thank you
onyx dew
I think when you divise 1 to sin its csc
finite birch
ngl im entering my second semester of uni engineering and i still don't know what csc means
odd edgeBOT
odd edgeBOT
Available help channel!
elfin wasp
Let ( a_k = \ln((2.8)^k) ). Find the minimal ( n ) such that ( \sum_{k=1}^n a_k > 190 ). (Give the minimal ( n ) that will definitely satisfy the inequality and support your claim with formal calculations)
clever fjordBOT
mjd
elfin wasp
I got an answer but I’m not sure if I did was right. I moved the k to the front even though I don’t think I can/should’ve. Is there a way to do this without moving k to the front or is what i did the correct way to do it?
elfin wasp
Idk I saw the double parenthesis in the ln argument and thought maybe it’s saying don’t move the k to the front
weary pelican
no, there's one set of parentheses around 2.8 to specify that the whole 2.8 is raised to the kth power, and there's another set of parentheses for the argument of ln
elfin wasp
Alright
Continuing on what I did
I moved the k to the front. We know that k summation is ((n+1)n)/2. So we get ln2,8((n^2+n)/2) > 190
Divide by ln, multiple by 2, you get n^2 + n> 380/ln2.8
I kind of got stuck here and used trial of error. Ik ln 2 is 0.7 and ln3 is 1.1, so I assume ln2.8 is 1
Approximately
And I subbed in values to get 19
weary pelican
ln(2.8) would almost be 1
ln(e) is 1
2.8 > e so ln(2.8) is slightly greater than 1
weary pelican
oops sorry
elfin wasp
Yea I got ur point it’s alright
Anyways do u think there’s a better way to do this without trial and error
Without a calculator
weary pelican
so there is a way to solve without trial and error
elfin wasp
I’m listening
weary pelican
well, you want to solve n^2 + n - 380/ln2.8 > 0
so start by solving n^2 + n - 380/ln2.8 = 0, and I hope you know how to do that
elfin wasp
Yea quadratic
weary pelican
yes
weary pelican
ofc not
you always will need a calculator, unless you approximate ln(2.8) with 1
and even then it stays messy
weary pelican
yes so that's the positive solution for n^2 + n - 380/ln2.8 = 0
but since we want the first integer such that n^2 + n - 380/ln2.8 > 0
we're gonna take the first integer after that real value (between 18 and 19)
weary pelican
we're gonna take the first integer after that real value (between 18 and 19)
yes that's what was said here
elfin wasp
Ahhh didn’t see that sorry
weary pelican
so 19 is the answer
elfin wasp
Checks out
weary pelican
ofc we must beware that we approximated ln(2.8) to 1, so if that was a bad approximation we might have gotten a bit off, like 20
thankfully it wasn't the case here
odd edgeBOT
odd edgeBOT
Available help channel!
warped barn
odd edgeBOT
warped barn
ancient idol
$h(g(x))=h(x)\Leftrightarrow g^{2}(x)=x^2\Leftrightarrow g(x)=x$ or $g(x) = -x$
clever fjordBOT
Alisia
ancient idol
now you have 2 cases
warped barn
how is $g^{2}(x)=x^2$?
clever fjordBOT
Jug
ancient idol
if $h(x) = h(y)$, then $x^2=y^2$ since $h(x)=x^2$
clever fjordBOT
Alisia
warped barn
but why not just $g(x)=x^2$
clever fjordBOT
Jug
ancient idol
am i right that hg(x) is composition of h and g?
warped barn
yes
ancient idol
ok so as given $h(x)=x^2$
clever fjordBOT
Alisia
ancient idol
then $h(g(x))=(g(x))^2$
clever fjordBOT
Alisia
warped barn
ohhhh
warped barn
$h(g(x))=h(x)\Leftrightarrow g^{2}(x)=x^2\Leftrightarrow g(x)=x$ or $g(x) = -x$
ok i get this
but it seems wrong because they gave $g(x)=x+5$ and we got $g(x)=x$
clever fjordBOT
Jug
ancient idol
it means x = x+5
warped barn
yeah but that dosen't make sense
ancient idol
but basically no solutions of this equality
warped barn
yeah
ancient idol
so what's the issue?
ancient idol
yes
slate moon
hello
slate moon
i have a question
I’m making homework and i have trouble finding this question
i think its D but im not sure
i wanna be sure can someone check?
drowsy root
!show
odd edgeBOT
Show your work, and if possible, explain where you are stuck.
drowsy root
the functions are really hard to read
slate moon
well my motivation would be c and d as vertical asymptotes and (x-b)^2 its even so it doesnt go trhough the zero point
(i can be wrong)
drowsy root
does it have a perforation there on the right?
slate moon
uh good question
i think it jusr show that is fx
but i see its confusing with the open circle
but i dont think its meant as perforation
slate moon
oh okay wd
drowsy root
i cant read what they say
slate moon
drowsy root
sure
slate moon
well honestly
this one is a bit trickier
and i cannot het the right answer right away
drowsy root
yes x=0 is one
what are the other ones
remember that f(x)=0 means the points where is hits the x-axis
slate moon
(x-b).(x-c)
drowsy root
(ignore the other parts for now)
drowsy root
how would we write the vertical asymptote
slate moon
(x-d)
slate moon
1/ (x-d)
drowsy root
what can you read in the graph for the horizontal asymptote?
slate moon
a/g
slate moon
(x-b).(x-c). (x-0) / (x-d)
drowsy root
we are still missing some parts right?
the horizontal asymptote is for when x->+/- inf
right now what happens when x->+/- inf?
drowsy root
but at the numerator we hve x^3 and in the denominator we have x^1
so it grows to infinity right?
how do we compensate this
(btw technically we dont need to do this, there is only one option left)
drowsy root
if we have x^3 in the top
and x in the bottom
we need another x^2 in the bottom
so that we have matching x^3 and x^3
slate moon
oh yeah i see
drowsy root
then when x grows really large it gets compensated and doesnt grow to infinity
slate moon
yes now it matches
drowsy root
!nosols
odd edgeBOT
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.
drowsy root
x-a isnt a root so this is not the function
with all this information
which is the only possible answer
drowsy root
that seems correct
slate moon
nice
drowsy root
yep
slate moon
thanks for the help
drowsy root
oh
wait
how many x's does the top and bottom have
thats H
"none of the above"
oh
you deleted your message
slate moon
like three x?
drowsy root
yes and the bottom as 4 of the right?
so x^3/x^4=1/x so if x->inf then f(x)->0=/=a/g
drowsy root
yes
but we needed to compensate for that at least
but in the possible answers
there isnt something that has roots x=0, x=b, x=c and vertical asymptote (inverse roots) x=d
also importatnt to note is that in the figure they also show -b and -c
so we know that -b and -c are not roots
drowsy root
that doesnt have to be the case
we can also add (x^2+something^2) to the top
this cant be a root because x^2 is always positive and something^2 is always positive
so we can also get x^5 in the numerator
but none of the functions given have that
slate moon
but why would we add like x^5
drowsy root
this function could be anything
we can add random stuff to the top and the bottom and still have it get these properties in the graph
in fact, there are infinitely many functions that we can write down
drowsy root
wdym?
slate moon
the key factor
because i also see many possibilities
and dont see why some functions cant work
drowsy root
the important part about these questions is finding the roots and the vertical asymptotes (inverse roots)
so you first find where f(x)=0
then where f(x)=+/- inf
and then usually you can already find the solution
if not, then look at x->inf or x->-inf and use the horizontal asymptotes
slate moon
the important part about these questions is finding the roots and the vertical a...
yes
drowsy root
most of the time you can remove half the functions just by seeing the roots
the another quarter with the vertical asymptotes
then perhaps a 50/50 with the horizontal asymptote
devout plinth
odd edgeBOT
odd summit
thats not in slope-intercept form
devout plinth
thats not in slope-intercept form
then what is it
cuz theres no y
so how would i do it
devout plinth
your equation is a vertical line
yea so im asking if x=9/5 is right
odd summit
yea so im asking if x=9/5 is right
no it is wrong
the question is wrong
oh wait
i just saw your image
it is some weird wording
but it might mean 5x-9=y
devout plinth
ok
odd summit
you can't write it in slope intercept form
if its standard form for linear equations
its in ax+by=c
odd summit
x int is 9/5 but but there is no y-int
devout plinth
so do i put undefined
odd edgeBOT
Available help channel!
willow shard
I have $S_n = a_1 + 2na_{n+1} + na_{n+2} $
odd edgeBOT
I have $S_n = a_1 + 2na_{n+1} + na_{n+2} $
willow shard
what is the limit of Sn?
where (a_n) is positive sequence that is decreasing with limit = 0
i'm sure the limit is a1 but how
woven kettle
any body explain the goat prblm
fallow thunder
$S_n = a1 + 2na{n+1} + na_{n+2}$
clever fjordBOT
LILY
willow shard
$S_n = a_1 + 2na_{n+1} + na_{n+2}$
clever fjordBOT
sheerxwood
willow shard
what is the limit of Sn?
willow shard
where (a_n) is positive sequence that is decreasing with limit = 0
given this
HELP PLS
why is the limit = a1
is 0 * infinity = 0 always ?
drowsy root
is 0 * infinity = 0 always ?
Depends
drowsy root
If you have an=1/n and bn=n, then lim an = 0 and lim bn = inf, but lim an*bn = 1
willow shard
where (a_n) is positive sequence that is decreasing with limit = 0
.
@drowsy root
GUSY PLS HELP
<
odd edgeBOT
@willow shard Has your question been resolved?
hidden salmon
d/dx {log_2√(x^2+1)}
odd edgeBOT
d/dx {log_2√(x^2+1)}
signal oar
d/dx {log_2√(x^2+1)}
$\frac{\dd}{\dd x} \Bigg[\log_2\Big(\sqrt{x^2 + 1}\Big)\Bigg]$
lime turret
$d/dx {log_2√(x^2+1)}$
signal oar
signal oar
How to start this
$\log_a(b) = \frac{\ln(b)}{\ln(a)}$.
Rewrite it like that
Since you most likely already know the derivative of ln
hidden salmon
Log_2 √(x^2+1)/2
hidden salmon
2?
hidden salmon
b is 1?
signal oar
No
hidden salmon
Ohh √(x^2+1)
hidden salmon
log√(x^2+1)/log2
signal oar
log√(x^2+1)/log2
Ok, so $\log_2\Big(\sqrt{x^2 + 1}\Big) = \frac{\ln\Big(\sqrt{x^2 + 1}\Big)}{\ln(2)}$, yeah
hidden salmon
I guess here you give them base value e
hidden salmon
U/v derivative?
signal oar
I guess here you give them base value e
This fact also holds with any base value, I did it with ln because you most likely already know that the derivative of ln(x) is 1/x, so we can use the chain rule here
cold urchin
,,\loglaws
clever fjordBOT
Pure
hidden salmon
1/ln2 {x/(x^2+1)
signal oar
Did you already take the derivative?
cold urchin
[\41{\ln2}\cd\4x{x^2+1}]
clever fjordBOT
Pure
cold urchin
This is the final answer yeah
signal oar
Did you already take the derivative?
Yeah
hidden salmon
So we can't solve the question directly with base 2
clever fjordBOT
Pure
cold urchin
Well you get that from the change of base thing
hidden salmon
logx/loga
1/loga {1/x}
cold urchin
That Kepe did
signal oar
clever fjordBOT
Pure
cold urchin
[\log(x^n)=n\log(x)]
clever fjordBOT
Pure
cold urchin
Any base
clever fjordBOT
Gitoo
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
signal oar
$\[\log(_a x^n)=?$
[\log_a\big(x^n\big) = n \cdot \log_a(x) = n \cdot \frac{\ln(x)}{\ln(a)}]
hidden salmon
Ohhh this is the actual mystery
signal oar
Or do you mean the derivative right away
hidden salmon
Yes i understand now how to change base and tackle power of log
Then derivative part is easy 😍
Thank you so much
.close
odd edgeBOT
signal oar
Then derivative part is easy 😍
The derivative part is usually just the chain rule
Also you could do it without the power of log property too
hidden salmon
Yes?
signal oar
Also you could do it without the power of log property too
It just makes it easier
hidden salmon
.reopen
odd edgeBOT
✅
signal oar
So I'd do it
hidden salmon
Yes as we did earlier?
signal oar
Yes as we did earlier?
Yes, I used it here
But you wouldn't have to use it, you could take the derivative of $\ln\Big(\sqrt{x^2 + 1}\Big)$ too, of course
hidden salmon
Yes true. I understand this question in many ways
signal oar
But you'd need to use chain rule twice
hidden salmon
You broke it totally 🥰
signal oar
And I think using power rule for logs is much faster
hidden salmon
.close
pearl patio
odd edgeBOT
pearl patio
Hi, so basically i did (100a + 10b + c) + (100c + 10a + b) + ... = 1332
so i had 222(a + b + c) = 1332
(a+b+c) = 6
i tried stars and bars on this but that's 5C2 and that's 10
does stars and bars not work because it's combinations but i need permutations here?
i did case works but i got 24 (but i think this is wrong):
600 -> 3 ways
510 -> 3! ways
420 or 411 -> 3! + 3
321 -> 3!
6 + 3(3!) = 24
kindred inlet
I think what you've calculated is not what the question is asking
pearl patio
I think what you've calculated is not what the question is asking
wdym?
i considered a digit (abc) that when permuted gives a sum of 1332
i think it's correct
kindred inlet
(100a + 10b + c) + (100c + 10a + b) + ... = X
it reads like they are asking how many distinct X can exist satisfying this condition
pearl patio
oh
pearl patio
well wait
pearl patio
i did case works but i got 24 (but i think this is wrong):
600 -> 3 ways
510 ->...
would you say 24 is correct?
kindred inlet
hmm since we cant have any of them as 0
pearl patio
hmm since we cant have any of them as 0 <:hmmCat:831508801646755850>
so :V how do i do it
is everything i did wrong?
kindred inlet
is everything i did wrong?
are we sure about what the question is asking
kindred inlet
(100a + 10b + c) + (100c + 10a + b) + ... = X
it reads like they are asking how ...
i think this makes most sense
pearl patio
are we sure about what the question is asking
i mean idk what the question is asking
but could you clarify
pearl patio
(a+b+c) = 6
if i have (a+b+c) = 6 then how many different integer possibilities do i have?
would the answer be 24 in this case?
just to make sure i'm doing combinatorics right
kindred inlet
if you say non zero I think so
pearl patio
if you say non zero I think so
i considered instances where they are 0 right?
kindred inlet
hmm I think we might be going on different directions, let me just say how I got 24
ancient idol
so i had 222(a + b + c) = 1332
right, but 1332 is just example
you should count different values of sum a+b+c possible
pearl patio
you should count different values of sum a+b+c possible
how would i do that?
kindred inlet
the least it could be is 6, and the maximum is 24(7+8+9)
pearl patio
the least it could be is 6, and the maximum is 24(7+8+9)
where did you get 24?
kindred inlet
where did you get 24?
by choosing the largest 3 digits
pearl patio
by choosing the largest 3 digits
ah i see
so the least is 6 and the max is 24
but wait where do i get 19 then
,calc 24 - 6 + 1
clever fjordBOT
Result:
19
oh
kindred inlet
yeah
and back to your question about number of positive solutions to a + b+c =6 I think 24 is right, but the method you showed didnt look right to me
it is equivalent to solving a+b+c=3, where they can be 0 (if im right) and that has 5C2 solutions
pearl patio
and back to your question about number of positive solutions to a + b+c =6 I thi...
5C3 is stars and bars right
but thta's combinations, no?
so 001 is counted only once?
when it should be counted three times?
pearl patio
1 combination
sorry that is contrived
here
a = 3, b = 0, and c = 0
in stars and bars this is counted as 1 right
kindred inlet
yes
pearl patio
but can't it be (3,0,0) or (0,3,0) or (0,0,3)
kindred inlet
yeah those are distinct and counted
pearl patio
u mean in stars and bars it accounts for all 3 tuples?
i thought it uses combination notation so it considers {3,0,0} as one
and order doesn't matter
kindred inlet
just because you calculate it using a nCk doesnt imply that
here it explains why it works for integer equations
kindred inlet
I tried typing it out but the itlaics and formatting
kindred inlet
and back to your question about number of positive solutions to a + b+c =6 I thi...
yeah thats why I said this
pearl patio
yeah but that's 5C2 again right?
i don't get it 😭
a + b + c = 3 wait
this one is 2C2 = 1
from stars and bars
5C2 is for a + b + c = 6 right?
kindred inlet
this one is 2C2 = 1
where did you get that from
kindred inlet
its N+K-1 choose K-1 for integer equations
um I might be mistaken
let's wait for someone more qualified to fix these things
kindred inlet
this one is 2C2 = 1
but 1 solution doesnt look right at all
since here we say a, b, c can be 0
pearl patio
i have to remind myself
of stars and bars
but i don't see how a + b + c = 3
would give 5C2 is all i meant
there's 5 items
2 dividers
5!/2! = 5C2
oh lol
whoops i was thinking about something else ig
okay then a + b + c = 6 would give 8!/2! -> 8C2 right?
clever fjordBOT
pearl patio
am i correct? @kindred inlet
kindred inlet
yeah a + b + c = 6 should have 8C2 non-negative solutions
pearl patio
yeah a + b + c = 6 should have 8C2 non-negative solutions
oh nice
where did i miss the 4 then?
just curious
in here
oh 330 and 222?
pearl patio
easy to find 😭 when i know the answer
i'm not sure how to do case work in a systematic way
pearl patio
330?
i'm guessing for a,b,c >=1
you had a redefinition right
a' = a - 1 >= 0
b' = b- 1 >= 0
c' = c- 1 >= 0
(a' + b' + c') + 3 = 6
(a' + b' + c') = 3
and this one was 5C2
without including 0's
kindred inlet
yes, that's what I did
pearl patio
yes, that's what I did
owhh thanks, one last question but for the actual question now
kindred inlet
altho my thought was more like "give 1 to everyone, allocate remaining 3" lol
pearl patio
222 ( a + b + c ) = X where X is distinct
(a + b + c) are all digits between 1 and 9 (inclusive)
a,b, and c can't be the same
lol 😭 idk if i'm dumb but why can't i just do 9C3
i get all subsets where each digit is unique then right?
kindred inlet
because for example 2, 5, 7 and 1, 6, 7 are distinct but have same a+b+c value
wait wrong example
pearl patio
ah cool
so i find min value of (a,b,c) and max value of (a,b,c)
how do i know each value is attained
in the interval
also not sure if it's a bother but can i get the intuition behind doing the min and max value thing
kindred inlet
its not really rigorous, but you can always +1 to some element in your set to get the next one (making sure not to duplicate)
pearl patio
it doesn't make sense to me but also does kinda