emm
#help-23
warm vault
no like apple or fish
spring horizon
oh ok
warm vault
or just a question mark
spring horizon
ohh
warm vault
this one
warm vault
just write :D instead pf D - A
spring horizon
can i use x
warm vault
yes of course
spring horizon
yup
warm vault
.
spring horizon
C - B?
warm vault
yes
spring horizon
ohh
warm vault
write down b instead of C - B
spring horizon
now what
warm vault
(C + x - B)/2?
write b in this equation
spring horizon
AD?
warm vault
no the lowercase a
warm vault
I think you do the algebra wrong
spring horizon
oh
warm vault
b + x/2 = a
oh sorry me
spring horizon
thats it?
spring horizon
ohh ok tyy
warm vault
you're welcome
spring horizon
.close
tranquil rover
Does anyone know what to write down for the reflective property. I'm confused on what to write down for it.
safe radishBOT
Does anyone know what to write down for the reflective property. I'm confused on...
misty bay
do you know what the reflexive property means
safe radishBOT
safe radishBOT
Available help channel!
dusty bramble
safe radishBOT
worthy hemlock
would the domain be (-9,-2)
No
dusty bramble
🥲
worthy hemlock
Recall that domain is all possible x values for that function
worthy hemlock
Yes
dusty bramble
damn i kinda felt like that with the range
worthy hemlock
Apply that logic to the range, what are the possible y values that function can have?
worthy hemlock
Not quite
worthy hemlock
Nope
worthy hemlock
What's the smallest y value you can have?
dusty bramble
10
worthy hemlock
How did you determine that the domain was (-inf, inf)?
dusty bramble
cus inf is all real numbers
dusty bramble
cuz the red part goes from both ends of the graph
worthy hemlock
Red part meaning the arrow?
dusty bramble
yes
worthy hemlock
So then what about the y values?
dusty bramble
10
worthy hemlock
Does the line ever reach 10?
dusty bramble
so 5?
worthy hemlock
Count again
safe radishBOT
@dusty bramble Has your question been resolved?
dusty bramble
the line reaches to -10
dusty bramble
the red line doesn’t reach -10?
dusty bramble
so 5?
dusty bramble
-3?
worthy hemlock
No
dusty bramble
-9
worthy hemlock
Where is -5 from?
dusty bramble
i mean 5
worthy hemlock
So how do you properly write that in interval notation?
dusty bramble
(5,-inf)
dusty bramble
(-inf,5)
dusty bramble
so just inf
dusty bramble
oh
worthy hemlock
Do you know the difference?
worthy hemlock
So is 5 included or not?
dusty bramble
so [-inf,5) ?
dusty bramble
thank you for help 🙂
safe radishBOT
@dusty bramble Has your question been resolved?
dusty bramble
.close
plain hedge
im stuck on this
viral magnet
This is an ACT question from a sample test. The answer is F. I do not quite understand how to get the answer. Thanks for the help in advance.
viral magnet
resolved
safe radishBOT
@viral magnet Has your question been resolved?
stark idol
dunno where to start
stark idol
<@&286206848099549185> *'
safe radishBOT
@stark idol Has your question been resolved?
stark idol
<@&286206848099549185>
flat frigateBOT
Saccharine
stark idol
but you can rearrange it to $|z|Re(z^2) = 175$
yeah i did that but not sure what to do after\
low flare
stark idol
<@&286206848099549185>
spiral crescent
safe radishBOT
spiral crescent
I understand why option B is the answer
And option A cant be the answer (assuming intersecting isn't talking of the scenario where they are the same line)
But what is stopping option C to be the answer
broken yew
Think of a case where C holds but they don't intersect
Try to construct an example
Just checking we're in 3D space right, else this makes little sense
spiral crescent
Just checking we're in 3D space right, else this makes little sense
Yes we are
What i mean to say is
Ignore the 0 on the lhs
broken yew
Without trying the question, I would recommend a diagram
to try thinking of a counter example
I wouldnt approach by algebra.
spiral crescent
But the rhs is the formula for the distance between two lines right, when the cross direction vectors is not 0
And if we look at the 2nd term in the dot product on the numerator, isnt that always 0?
b cross c would be perpendicular to c, and c cross a would also be perpendicular to c, so wouldnt the angle between these two vectors be 0 or 180 in both cases the cross is 0?
Oh wait but then the thing that were dividing by is also 0 wtf
spiral crescent
I wouldnt approach by algebra.
My teacher gave us a flowchart for these questions, i assume the proofs do come from a diagram but due to limited time we have to memorize the formulas
spiral crescent
broken yew
breh wut
spiral crescent
For context b1 and b2 are direction vectors of the lines
broken yew
sounds crazy to memorize this
spiral crescent
sounds crazy to memorize this
Its actually very easy as compared to the flowchart for cramers rule
broken yew
You can memorize this and fail to understand what is going on geometrically
spiral crescent
You can memorize this and fail to understand what is going on geometrically
I do have the proofs written down as well
safe radishBOT
@spiral crescent Has your question been resolved?
safe radishBOT
@spiral crescent Has your question been resolved?
spiral crescent
.close
silent scaffold
given 3 real positive numbers a,b,c where a+b+c = 1: how do i find the max value for abc
safe radishBOT
given 3 real positive numbers a,b,c where a+b+c = 1: how do i find the max value...
glass carbon
yes it is
safe radishBOT
silent scaffold
.reopen
safe radishBOT
✅
silent scaffold
its the same as squares
like
if you have a+b=c
and the max value for a*b is just a=b
idk the proof
lean otter
yeah i know that its "like" that but can you prove
timid pasture
so, in general for x1 + x2 + x3 + ... + xn =1, the max value is 1/n ?
lean otter
so, in general for x1 + x2 + x3 + ... + xn =1, the max value is 1/n ?
1/n
cuz you know n is constant
glass carbon
you can prove it using AM-GM
glass carbon
$$\frac{a+b+c}{3} \geq \sqrt[3]{abc} \iff \frac{1}{3} \geq \sqrt[3]{abc} \iff \frac{1}{27} \geq abc$$
flat frigateBOT
Modus
silent scaffold
ill try to
oh i dont know how to proof it(i never tried to think about why it is so)
glass carbon
real mean would use multiple variable calculus
silent scaffold
am gm is just arithmetic mean geometrical mean right?
glass carbon
am gm is just arithmetic mean geometrical mean right?
yep
glass carbon
often used to prove different inequalities
silent scaffold
oh ok
glass carbon
in fact not only AM-GM but also square mean and harmonic mean, but AM-GM is used most frequent
silent scaffold
hmm
do you know any videos/blogs or anything like that you could recommened which explains am-gm?
glass carbon
youtube I guess
echo sequoia
So, variance is the average of the difference from the mean, right?
echo sequoia
I get this equation: V=Sum(x_i - x_avg)^2/n(x)
where n(x) is the number of samples
but my book says it is V=Sum(x-mu)*p(x)
echo sequoia
oops typo
But with the square included, is there really a difference between the two outcomes?
peak estuary
well one is if you have a dataset and calculate the variance of that
the other is if you have an abstract random variable and calculate the variance of that
or I suppose you can also see the first as having a random variable where all results are equally likely
echo sequoia
well I get the same mu considering a random variable vs considering equal likelihood.
I'm using the number of heads when flipping three coins as an example.
(x) | 0 | 1 | 2 | 3
p(x)|1/8|3/8|3/8|1/8
avg = (0+1+2+3)/4 = 3/2
mu = (0 * 1/8) + (1 * 3/8) + (2 * 3/8) + (3 * 1/8) = 3/2
V(x) (as a sample) = [(9/4)+(1/4)+(1/4)+(9/4)]/4=20/16
V(x) (as a random variable) = [9/4*1/8]+[1/4*3/8]+[1/4*3/8]+[9/4*1/8]= 12/16
Is my work accurate, those are just discrepancies when considering a sample vs a random variable?
or am I doing something wrong?
peak estuary
yes the first would be if you do the experiment 4 times and get the results 0,1,2,3 once each
and the other you can do even before actually doing the experiment
echo sequoia
Okay... Then I am on the wrong track to understanding. My ultimate goal here is to understand how to show this...
peak estuary
well they both satisfy that
echo sequoia
But to do that, I need to understand the V(x) equation well enough to use V(aX+b)
echo sequoia
We're doing random variables right nowj
which the only equation I have for the random variable variance is this:
I had work but it got so convoluted with summation notation everywhere that I was unable to see the proof tree through the summation forest.
peak estuary
It's just one sum
echo sequoia
So this is not right?
echo sequoia
But how do you evaluate the inner sum separately in a general proof?
peak estuary
ok, so we have $X$ which is a random variable and know that $\mu_X = \sum x p(x)$ is the expected value
flat frigateBOT
Denascite
peak estuary
we now define the random variable $Y=aX+b$ and want to show that $V(Y) = a^2 V(X)$. as a first step, let us calculate $\mu_Y$ in terms of $\mu_X$
flat frigateBOT
Denascite
peak estuary
what is mu_Y
echo sequoia
okay...I will try this then using the $ notation there.
peak estuary
that's for latex and just there so that it looks nicer
peak estuary
nearly
flat frigateBOT
BlewiiQ
peak estuary
small x but yes
flat frigateBOT
BlewiiQ
echo sequoia
?
peak estuary
that will come but a bit later
peak estuary
**BlewiiQ**
lets stick with this for a second
can you multiply out the bracket and separate the sum?
echo sequoia
$\mu_Y = \sum(ax\cdot p(x) + b\cdot p(x))$
flat frigateBOT
BlewiiQ
echo sequoia
like that?
echo sequoia
well a and b are constants, right? but x and p(x) are not, so there's something there to be done Im sure.
peak estuary
yes
echo sequoia
I don't remember rules for summation notation, but I think there is one where a constant times a variable in a sum is the same as the variable, summed, times the constant?
i.e. $\sum ax = a\sum x$
flat frigateBOT
BlewiiQ
echo sequoia
?
flat frigateBOT
BlewiiQ
peak estuary
yes
peak estuary
**BlewiiQ**
and now this twice
flat frigateBOT
BlewiiQ
echo sequoia
mu, right?
echo sequoia
ooooh. I didn't even think about that, wouldn't it be 1?
flat frigateBOT
Denascite
echo sequoia
makes sense there, and so now we would...
peak estuary
now we turn our attention to V(Y)
echo sequoia
$V(Y)= \sum (Y - \mu_Y)^2\cdot p(x)$
flat frigateBOT
BlewiiQ
echo sequoia
?
echo sequoia
which we fill in the variables Y and mu_Y?
peak estuary
yes
echo sequoia
$V(Y) = \sum [(ax+b) - (a\mu_X + b)]^2 \cdot p(x)$
flat frigateBOT
BlewiiQ
peak estuary
yes
echo sequoia
and the b's would cancel
peak estuary
yes
flat frigateBOT
BlewiiQ
flat frigateBOT
BlewiiQ
echo sequoia
?
peak estuary
yes but you don't need to multiply it out
echo sequoia
you don't?
peak estuary
but keep going, you'll get the same result
flat frigateBOT
BlewiiQ
peak estuary
yes
echo sequoia
which should be the same as $a^2\sum [x-\mu_X]^2\cdot p(x)$
flat frigateBOT
BlewiiQ
peak estuary
yes
echo sequoia
Thus it's proven there, right? because the sum x - mu_X squared times p(x) is V(x).
peak estuary
yes
echo sequoia
awesome. You're a time/money/life saver, Denascite!
peak estuary
most of this you did yourself
try proving it for the other variance formula from earlier
it follows roughly the same steps
just as an exercise
echo sequoia
Well, I didn't understand that you could just keep mu as mu_x and mu_y within the first summation.
peak estuary
but then you got it and adapted really quickly and were able to do all the steps
echo sequoia
I wouldn't use p(x), and n(x) would be outside of the sum itself, right? Therefore I just have to show that the average of an altered sample is altered in the same way
peak estuary
yes
you can also see it as just setting p(x) = 1/n(x) and then pulling it out everywhere but maybe it's a good idea to do the whole thing by hand a second time
never hurts
echo sequoia
Okay then, you're a significantly better instructor than my Professor in the course, I wish my University kept you on retainer instead of him lol.
echo sequoia
He doesn't reply to my one-on-one emails >.>
peak estuary
professors have a lot to do
sadly
just not that much time left for all the teaching stuff
echo sequoia
well, he's not as good at doing it as my other professor. Because the other one can reply to all of my emails and still get it all done. And it's an online course btw.
peak estuary
maybe your other prof is just really good
echo sequoia
Anywho, I'll call this done then, and let you get back to life. Thanks a ton, regardless of the quality of my professors lol.
peak estuary
👍
knotty gate
Can somone explain me how to do this ?
safe radishBOT
Can somone explain me how to do this ?
@knotty gate Has your question been resolved?
safe radishBOT
@knotty gate Has your question been resolved?
safe radishBOT
@knotty gate Has your question been resolved?
safe radishBOT
@knotty gate Has your question been resolved?
safe radishBOT
@knotty gate Has your question been resolved?
plucky elk
Can somone explain me how to do this ?
Translate if you want help
scenic valley
People I hope y’all having a nice day and fun and enjoyment learning math! 😃
lean otter
what did i do wrong?
safe radishBOT
what did i do wrong?
lean otter
integration by parts two times
plucky elk
!show
safe radishBOT
Show your work, and if possible, explain where you are stuck.
tawdry plover
sign of the sinx cos x looks wrong
i think sin^2 is 1/2-1/2cos2x?
or 1/2 - sin cos
@lean otter
safe radishBOT
safe radishBOT
Available help channel!
dense token
safe radishBOT
pseudo scroll
Use the general tan 2x identity
$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$
flat frigateBOT
NEONPerseus
pseudo scroll
I hope I got that right
pseudo scroll
That works
safe radishBOT
@dense token Has your question been resolved?
dense token
alright thanks
west rampart
I need some help with exercise d)
safe radishBOT
I need some help with exercise d)
west rampart
d) Construct the point of intersection T of the straight line g with the yz-plane in an oblique drawing.
(I hope my translation is correct)
I also have the solutions:
safe radishBOT
@west rampart Has your question been resolved?
west rampart
<@&286206848099549185>
west rampart
<@&286206848099549185>
west rampart
I think so
frank glen
Looks like your solution is correct
safe radishBOT
@west rampart Has your question been resolved?
west rampart
Looks like your solution is correct
I know it's correct, but I don't understand why
frank glen
Hmm
The yz plane appears as a line when you look at it on the xy plane right?
Also you find the projection of the line g on the xy plane
Then you find the intersection of the yz plane and the projection of line g
That's how you find the x and y coordinates of the point T
Then you find the z coordinate by looking at the xz or yz plane
young nexus
Then you find the z coordinate by looking at the xz or yz plane
Task a) is to find the equation for the line. if someone has managed this setting x = 0 in this equation would give the y and z coordinates from T. So what is the advantage of this graphical method?
frank glen
Hmm
Is this analytical geometry or descriptive geometry?
I'm speaking on the basis of using Descriptive geometry
young nexus
task a) is to determine the equation for the a plane and a line.
task b) is to determine if there is a common point
task c) is to calculate an angle.
and then comes task d).
young nexus
yes of course. but a) - c) are calculations, d) should then be done graphically - as the suggest solutions show. I understand why someone has trouble to understand this switch. and at the end corinne is asking for an explanation for this.
frank glen
Okay
When dealing with planes, it is best to look at them from an angle that makes them look like lines
Or on another angle that makes appear in their true shape
Now the intersection of a line and a plane can happen anywhere on the plane, so to get a better image or where the line can intersect the plane, you look at the plane of interest as a line
The yz plane that we want to intersect with a line looks like a line when we view it from the top
Like so
We do this so that we can turn the 3D problem into a 2D problem
This in turn makes our job easier
What remains is the line g
Which still appears as a line
So we can find the point of intersection between it and the plane yz (which appears as another line)
The line g and the yz plane now intersect at a point T from the view
We now have the y coordinate and the x coordinate
Let's find the z coordinate
By looking at another view where we graphed the z and y coordinates, we can find the z coordinate of the point T
While having its x and y coordinates
Thus making us find the point with ease
Here we viewed plane yz as its true size
And thus makes it easy to find what we want from it
The z coordinate
young nexus
thanks (for me).
safe radishBOT
@west rampart Has your question been resolved?
frank glen
@west rampart is the explanation good?
safe radishBOT
west rampart
<@1049708256832000060> is the explanation good?
Yes thank u:)
safe radishBOT
Available help channel!
lean otter
let y = 2+3i and z = 1+i
y*+z*=*(y+z)
can I factor the conjugate out like above
safe radishBOT
let y = 2+3i and z = 1+i
`y*+z*=*(y+z)`
can I factor the conjugate out like abov...
thin bridge
what's the the * after the =
glass carbon
let: y = a + bi, z = c + di
y* = a - bi
z* = c - di
y* + z* = (a+c)-(b+d)i
(y+z)* = (a+bi + c+di)* = ((a+c) + (b+d)i)* = (a+c) - (b+d)i
hence
sum of conjugates is conjugate of sum
safe radishBOT
@lean otter Has your question been resolved?
thorny root
can someone help me with part (d)?
thorny root
<@&286206848099549185>
safe radishBOT
@thorny root Has your question been resolved?
tough jackal
So we need to find the tangent vector, r'(t) and evaluate it at the point given. We find the value of t from setting the z component equal to the height (ie 18=2t). Evaluate the tangent and position vector at this point and make new equations.
x=at+x_0
y=bt+y_0
z=ct+z_0
where a,b,c are found from evaluating the tangent vector, and x_0, y_0, z_0 is found from evaluating the position vector
safe radishBOT
@thorny root Has your question been resolved?
lean otter
whats the contrapositive of this statement
safe radishBOT
whats the contrapositive of this statement
lean otter
If $a_n \rightarrow L_1$ & $a_n \rightarrow L_2$ then $L_1 = L_2$
flat frigateBOT
! matthewzz
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lean otter
Whats the contrapositive
is it
L1 != L2, then an doesnt converge to L1 or an doesnt coverge to L2
lunar owl
<@&286206848099549185> can someone help me question 1 part a?
safe radishBOT
<@&286206848099549185> can someone help me question 1 part a?
Please read #❓how-to-get-help
empty gyro
Zac my savior
It's pronounced Sac
lean otter
ya last quarter
empty gyro
(Or Sigmaac)
final halo
a mystery of the ages
empty gyro
Any deltas?
final halo
yes, disgusting
lean otter
a couple deltas so far
empty gyro
Reasonable
lean otter
i was sick the day we went over continuous definition
final halo
maybe you long to see a basis again one day
empty gyro
Topology will be more fun
lean otter
i want to go into stochastic calc
or something thats applied
no more epsilons or i throw up
final halo
you'd be hard pressed to totally avoid them in anything with the word stochastic in it lol
lean otter
**! matthewzz**
Compile Error! Click the <:errors:504266556054962196> reaction f...
back to this
lean otter
L1 != L2, then an doesnt converge to L1 or an doesnt coverge to L2
or rather this
lemme think
final halo
what you said was correct
if you remember that logical OR means that neither things can happen too
empty gyro
Depends on the problem imo
final halo
i think you do contradiction iirc
assume L1 != L2
show L1 = L2 using the limits
arrive at some contradiction
flat frigateBOT
! matthewzz
lean otter
or am i going the wrong direction
empty gyro
Ah are you trying to prove that a limit is unique?
lean otter
yes
empty gyro
I have this proof written in old notes somewhere. Let me refresh myself real quick
Iirc I did not use contradiction. But it's totally valid to do so
Okay
Consider this
L1!=L2
flat frigateBOT
! matthewzz
lean otter
I suppose L1 != L2
lean otter
**! matthewzz**
then im thinking theres some way to manipulate this to L1 <= L2
and then do the same thing reverse
L2 <= L1
empty gyro
$|a_{n}-L_{1}| - |a_{n}-L_{2}| \leq 0$
This was different from how I did it. Not saying this approach is right or wrong. I just haven't looked into it so am unsure
flat frigateBOT
! matthewzz
lean otter
$|a{n}-L{1} - a{n}-L{2}| \leq 0$
flat frigateBOT
! matthewzz
lean otter
$|-L{1} -L{2}| \leq 0$
final halo
definitely not absolute values dont work like that
flat frigateBOT
! matthewzz
lean otter
but by triangle inequality?
final halo
where are you applying triangle inequality
final halo
u should recall what triangle inequality says
flat frigateBOT
! matthewzz
lean otter
$|a+b| \leq |a| + |b|$
flat frigateBOT
! matthewzz
lean otter
is the triangle equality
final halo
+
flat frigateBOT
! matthewzz
lean otter
**! matthewzz**
should i go this direction?
@final halo
or can i just drop the absolute value 💀
final halo
absolutely (heh) do not drop absolute value
you probs will have to use triangle ineq at some point but i think youre going down the slightly wrong path
try writing out the definitions of the limits and choosing your epsilons carefully
lean otter
$\text{Suppose } a_{n} \text{ converges to } L_{1} \text{ and } L_{2}$
$\text{The for all } \epsilon> \text{0, there exists N such that } \forall n \geq N, |a_{n} - L_{1}| \leq \epsilon \text{ and } |a_{n} - L_{2}|\leq \epsilon$.
lean otter
try writing out the definitions of the limits and choosing your epsilons careful...
my brain is telling me to subtract them
I dont need to define the epsilons/ N's differently because theyre from the same sequence right?
final halo
not sure what you mean by "from an"
final halo
if L1 = L2 then sure, but thats what youre trying to prove
at the moment you just have two unrelated limits
no reason for the ns to be the same
lean otter
can we say anything about the epsilon's
final halo
you can choose epsilon to be whatever you want
lean otter
Ok
So
Assuming an converges to L1 and L2
$\text{Then for all } \hat\epsilon> \text{0, there exists }\hat N \text{ such that } \forall n \geq \hat N, |a{n} - L{1}| \leq \hat\epsilon$
flat frigateBOT
! matthewzz
lean otter
$\text{Then for all } \bar{\epsilon}> \text{0, there exists }\bar{N} \text{ such that } \forall n \geq \bar{N}, |a{n} - L{2}| \leq \bar{\epsilon}$
flat frigateBOT
! matthewzz
dusk forum
So far I am having lots of trouble formulating the equations in order to solve
Instead of using system of inequalities
I used this equation to find both the x's = x(50-2x)=100
safe radishBOT
@dusk forum Has your question been resolved?
grave berry
So far I am having lots of trouble formulating the equations in order to solve
...
Alright, so what would the steps of this problem be?
What do you think you need to know and in what order?
Because the equations you need to maximize and satisfy will fall out of that
dusk forum
So far this is how I set it up:
if the width of the garden is x
then in the perimeter, the width makes up 2x(because the two sides)
so, the length is 50-2x
width is x
so multiply them together to get the area
area
we know that the area has to be 100 or less
Since it's asking for dimensions I set it to (50-2x)(x) = 100
From solving that I get either 22.8 or 2.2 as x
Since the length has to be less than 40
because the house is only 40 ft wide
so 50-2x has to be less than 40
So I used the other x value
and substituted the values back in to get
the dimensions to be 4.4*22.8
Are the equations and the end result right?
grave berry
There's one more constraint
dusk forum
Is it that the area has to be 100
grave berry
The area has to be 100 or less, the maximum width of the bed is 40 feet, and I think you successfully implemented it, but you only have 50 feet of perimeter fence
fresh birch
Is my proof right here?
safe radishBOT
Is my proof right here?
wheat cave
,rccw
flat frigateBOT
wheat cave
looks good
wheat cave
,rccw
flat frigateBOT
wheat cave
I think the formal theorem of congruency gives 5/3=x/4
wheat cave
Hi, does this look right? <@266573406919589888>
but otherwise its right, the result is the same
fresh birch
but otherwise its right, the result is the same
Yeah, so my proportions are still right tho, right?
Bc I matched up the corresponding sides and all
wheat cave
yes, x=20/3
wheat cave
yes
fresh birch
Lemme solve them rq
wheat cave
Lemme solve them rq
actually, please make new help channels, I have to head to bed soon
fresh birch
actually, please make new help channels, I have to head to bed soon
Kk thx for ur help
worthy hemlock
.close
simple heart
Hi I am trying to do the exercize 4.X of the PDF and I have a question
plucky elk
Hi I am trying to do the exercize 4.X of the PDF and I have a question
Don't make people download files to help you. Just screenshot and upload
simple heart
OK I'm doing it right now sorry
tall bough
Yeah my phone just got hacked 
simple heart
So I have defined the relative error as the intagral of (ax+b-log_2(x))^2 for x in ]gamma, 256[
And since I can express a primitive of the functions in the integral, I can give an expression of the integral
It gives me a function of the form f_gamma(a,b)
And I have tried to calculate the critical points of f_gamma to determine a (a, b) that minimizes the integral
But the critical point I found is not a minimum
Does anyone have an other idea to answer the questions of the exercize 4.X?
safe radishBOT
@simple heart Has your question been resolved?
safe radishBOT
@simple heart Has your question been resolved?
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@simple heart Has your question been resolved?
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@simple heart Has your question been resolved?
lean otter
safe radishBOT
lean otter
25c, please tag me when responding
icy lance
your question.... doesnt ask anything - are you asking for intersections? Reflections in lines?? @lean otter
icy lance
should be 2-y not 4-y? im not familiar with the shell method but im not sure where that 4 came from
@lean otter
safe radishBOT
@lean otter Has your question been resolved?
lean otter
.close
lean otter
25c, use shell method to rotate around the x-axiss
safe radishBOT
25c, use shell method to rotate around the x-axiss
lean otter
@stray socket sorry i didnt see ur message
lean otter
Then what's your radius
y?
lean otter
What's your height
sqrty - y +2
lean otter
oh sorry lol
lean otter
we could do washer
im sure the teacher wouldnt mind
let me try doing washer myself
wait so where is the inner r and outer r in this example
stray socket
Ye
safe radishBOT
@lean otter Has your question been resolved?
hexed fox
@lean otter please do .close
stray socket
Okay at this point I'm adamant you're just here to Shitpost
safe radishBOT
@lean otter Has your question been resolved?
sacred vale
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sacred vale
I did this
and partail fracs and got a = 1/2
and b = -1/8
is that so far correct?
devout shale
You can double check your partial fractions by making them have common denominators, taking their sum, and seeing if it is equivalent to your original fraction
sacred vale
You can double check your partial fractions by making them have common denominat...
this is my work
this my work
devout shale
okay
so if you want to know if that partial fraction decomposition is correct, substitute A and B in as what you solved for them and see if the equation is equivalent to what you started with
sacred vale
just the remaining integral
devout shale
okay
do you see any way of rewriting that first integral so you can use the reverse power rule?
and what might you use for substitution in the second integral? u=...?
sacred vale
do you see any way of rewriting that first integral so you can use the reverse p...
i just subs
the first one
nvm
devout shale
not quite, but close
you incorrectly rewrote 1/sqrt(x) as x^(1/2)
it is actually x^(-1/2)
so the integral is evaluated differently
don't forget the negative
devout shale
not quite
1/sqrt(x) is equivalent to x^(-1/2)
rewrite it like that and then try to use the reverse power rule
devout shale
there you go! so far so good
sacred vale
how would we subs sqrtx+1 tho
devout shale
that one is a bit trickier, it involves some algebra, but use the substitution u=sqrt(x)+2 for the second integral and you should be able to work something out that is more doable
sacred vale
that one is a bit trickier, it involves some algebra, but use the substitution u...
ye
its hard
idk how
devout shale
you forgot the dx
on the right side
add the dx, then write the fraction with the sqrt x on the bottom instead of as a power
solve for dx by itself
see if you can then use your defined u, to get rid of the x that remains when you solve for xd
dx*
devout shale
yes now solve for dx
dx=2sqrt(x) * du
and then remember what you defined u as, u=sqrt(x)+2
so u-2=sqrt(x)
and then 2(u-2)=2sqrt(x)
so
sacred vale
devout shale
yes now solve for dx
go step by step through what I did from here on, on your paper
I think you will be able to follow it that way
sacred vale
oh i c
sacred vale
go step by step through what I did from here on, on your paper
]
so this would be integral
devout shale
I believe so
safe radishBOT
@sacred vale Has your question been resolved?
safe radishBOT
sacred vale
.reopen
safe radishBOT
✅
safe radishBOT
@sacred vale Has your question been resolved?
devout shale
seems right to me
sacred vale
seems right to me
shows up wrong
sacred vale
I miss webwork
lol
devout shale
but anyways
sacred vale
what would i ahve done wrong here
devout shale
well, your partial fractions probably were incorrect
but there is a much easier way to do this integral
so let's just restart
and I will give you the idea
pull out 1/4 from the original integrand
and then inside the integral, take out a factor of 1/sqrt(x)
then there should be a substitution you can make, similar to the one we made earlier, where u=sqrt(x)+2
from there, solve for dx... replace the x's using your definition of u... (just like earlier).... and you will be left with a very nice and simple integral in terms of u
let me know once you have done that, and I will let you know if you got it correct
sacred vale
well, your partial fractions probably were incorrect
we have to do it by partial fractions
this is correct?
devout shale
can I see the entire question?
sacred vale
devout shale
yes that is correct
devout shale
I'd like to point out that this homework is online, so I am assuming only your final answer is important not the exact process
sacred vale
these were my partial fractions
sacred vale
I'd like to point out that this homework is online, so I am assuming only your f...
yes but i do it for practise
devout shale
those are not correct
sacred vale
how so
devout shale
this is not an integral you would use partial fractions for, so it is poor practice
they are incorrect because sqrtx and sqrtx+2 are not linear factors
so you do not factorize this way
devout shale
no
if you want to do partial fractions for this question
which I strongly do not recommend
you must first multiply by the conjugate
of the denominator
and then factor both the top and the bottom
in a typical way
which will leave you with two linear factors
and then you can perform your decomposition
if you wish
but, this integral would be much easier to do without
spiral crescent
is in partial fractions
Maybe do a x=t^2 sub first?
sacred vale
well, your partial fractions probably were incorrect
ok
i did it this way
i got this
devout shale
yeah I see it already
sacred vale
oh i did
devout shale
when you pulled out the 4
okay, well if you have a 2 instead of the 4 I believe your work is correct! nice job
sacred vale
ye
devout shale
((e^x)^2+8(e^x)+15)
a polynomial where your x is e^x
can you do partial fracs with that information
sacred vale
ah ic
sacred vale
can you do partial fracs with that information
devout shale
Honestly, if that is allowed then yes that would be right
I have never done partial fractions in this way
XD
but I expect if you are expected to do them like this, this would be how
sacred vale
but I expect if you are expected to do them like this, this would be how
ok for a i got
A= -12.25
B= 10/8
devout shale
no
sacred vale
what