green lily
ok i got it
#help-28
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next tulip
Can someone help me get through this step by step? I have to write it out but im stuck on what to do to solve
ember shadow
there are standard formulas for this
next tulip
well im asking here
ember shadow
volume of a cylinder is $\pi r^2 h$. surface area is $2\pi r^2 + 2\pi r h$
glossy valveBOT
cwatson
next tulip
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frigid kelp
I am trying to convert double integrals from Cartesian to polar. I understand the x=rcosθ and y=rsinθ.
Still I keep making mistakes somewhere.
devout inlet
A lot of effort to decipher.
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@frigid kelp Has your question been resolved?
frigid kelp
Apologies. Handwriting is not the best
The rectangular limits are
1<x<sqrt3
1<y<x
Which I need to convert into polar
However I simply don’t know how to
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mystic rune
Will send a picture of my work so far in a moment, but I tried doing this problem by manipulating the inequality abs(fn(x) - 0) to show that it is less than all epsilon. I got stuck at finding an n s.t (1/2)^n < epsilon is always true so I'm not sure if I'm using the right method
mystic rune
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mystic rune
<@&286206848099549185>
If someone does start helping me, could you ping me? I'm doing other assignments so I may not see your message immediately 😅
Sorry, I didn't realize university questions should go in their respective channels. I'll close the room ✨
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torn jolt
torn jolt
help
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gentle anchor
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gentle anchor
I did all the working im just not sure if i got the correct answer
vast fossil
It's correct
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torn jolt
An airplane lifts of 600 ft from the base of a cliff which is 125 ft tall. What is the minimum angle it can climb to clear the cliff?
torn jolt
How would I sketch this?
The wording feels weird...
it would be a triangle
draw the ground first
then the plane and the bottom of the cliff
then the top of the cliff
Like that?
yeah
Then can I do the tan(x) = 125/600
To solve for the angle
yup
Ok, can I tell you what I get?
,w arctan(125/600) to degrees
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twin matrix
71,65,61,55,45 what will be the next two number of the series, pls explain
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sleek escarp
How do I create a formula for a graph out of nothing? Like I have a description of it and a drawing but how do I get the formula
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How do I create a formula for a graph out of nothing? Like I have a description ...
sleek escarp
Bruh well idk how do close the other one
Ok there
Happy now bot?
So the discription of the graph I want to create is:
exponentially increasing up to a point then linearly continue then exponentially decrease up to a point and keep decreasing linearly infill 0 and then keep constant (cuz 0 is the lowest you can go)
It would then kind of look like this
And idk what exact values I want yet for like the linear parts and the “up to a point” values
So like can I make those configurable? Like diff variables
<@&286206848099549185>
Nobody knows?
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grizzled ridge
mighty stag
An exponential function is given by f(x)=20∙1.15^x) Sketch the graph in the range −5≤𝑥≤10 How do I sketch this?
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An exponential function is given by f(x)=20∙1.15^x) Sketch the graph in the rang...
grave nymph
I'm trying to solve this proof but I'm confused on which of these is the right option?
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signal solar
https://youtu.be/3s1WYUWYEKU?t=316
got completely lost once the integration part hit
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https://youtu.be/3s1WYUWYEKU?t=316
got completely lost once the integration part...
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silver pasture
i think this question is really silly but still cant figure it out myself
strong crystal
e^ln(y)=e^a, well e^ln(y)=y=ln(e^y)
y=e^a
e^ln(y)=y=ln(e^y) this is the definition of ln, btw
twin herald
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twin herald
what did I mess up
I keep getting -2/(n*Pi)
as a result
but it is wrong
<@&286206848099549185>
fossil stump
@twin herald Maybe check with an integral calculator? I can't look through that all at the moment
rain plinth
by that definition of f(x), f(5) should not be defined
twin herald
rain plinth
by that definition of f(x), f(5) should not be defined
but anyway
if you happen to know the fourier series for square wave you can graph it and then adjust stuff until it's the function
but ig that would be cheating
rain plinth
you integrate the function times sin(npix/2)
so that's int 0 to 2 of 2*sin + int 2 to 4 of 4*sin
and that gets the right answer
twin herald
so that's int 0 to 2 of 2*sin + int 2 to 4 of 4\*sin
huh wdym
rain plinth
$\frac{1}{L}\int_{0}^{4}f(x)*\sin(\frac{n\pi*x}{2})dx$
glossy valveBOT
zfnQRZJT
rain plinth
this
twin herald
so what is it exactly
this one right
nvm I figured it out
still didn't get it
I got -4/(n*Pi)
this time
rain plinth
I got -4/(n*Pi)
yeah that's right
\
rain plinth
well tbf
it does say 2k - 1, not n
ebon grove
any idea what could come next?
2 2 2 4 24 576 ?
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any idea what could come next?
2 2 2 4 24 576 ?
ebon grove
whats the most obvious pattern
rain plinth
is there a logic for this?
yeah you can construct a polynomial that fits any finite sequence
ebon grove
I know that
rain plinth
which means it would fit 2 2 2 4 24 576 0
wouldn't be the simplest answer but it would be a, a pattern
ebon grove
I think its supposed to something of the form
a_(n+1) = f(a_n)
rain plinth
I think its supposed to something of the form
`a_(n+1) = f(a_n)`
well that's clearly not possible
because 2 goes to 2 but then later 2 goes to 4
fails vertical line test
ebon grove
hmm, then maybe
a_(n+1) = f(a_n, n)
rain plinth
that's possible
ebon grove
1 1 2 6 24
this seems familiar
factorial
0! 1! 2! 3! 4!
2*0! = 2
2*1! = 2
2*2! = 4
4*3! = 24
24*4! = 576
576*5! = 69120
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ashen pewter
how can i solve this?
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how can i solve this?
ashen pewter
i have no idea how or where to start
thorny horizon
consider the volume of the tank to be V
thorny horizon
call the rate at which one pipe operates R
and the rate that the other pipe operates at r
ashen pewter
yeah
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@ashen pewter Has your question been resolved?
torn jolt
what is "theta" and "phi" here... i meant how did they arrive at conclusion that if we take tan (theta) or tan (phi) we will get this relation
spice orchid
Are you asking why θ = α2-α1?
torn jolt
no
spice orchid
What exactly are you asking then
grave elm
are you asking about this?
torn jolt
are you asking about this?
no
i mean why we get this anle b/w two lines relation when we find tan (theta) here??
spice orchid
The steps are laid out there for you, which do you not understand
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spice orchid
I don't think it was stupid it was just hard to figure out exactly what you meant
torn jolt
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spice orchid
Are you looking for motivation as to why you would take tan of θ?
hollow zenith
How do I solve this?
$5>\sqrt{(x+4)(x^2+4)}$
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How do I solve this?
$5>\sqrt{(x+4)(x^2+4)}$
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tobi.
bitter plover
you could start by getting rid of the square root
hollow zenith
but Im not allowed to sqaure both sides right?
bitter plover
you are
hollow zenith
no thats not an equivalent tranformation
tropic canopy
why not?
hollow zenith
I thought im not allowed to do that
tropic canopy
you are
hollow zenith
maybe im wrong
torn jolt
why note mate
hollow zenith
ah thx then I can solve this
bitter plover
the sign only flips if you multiply or divide by a negative number
hollow zenith
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vale karma
In this exercise, once I have found "w", how can I obtain the ln(w)^1/2?
vale karma
I have to use this formula
vale karma
In this exercise, once I have found "w", how can I obtain the ln(w)^1/2?
This is the answer
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@vale karma Has your question been resolved?
torn jolt
<@&286206848099549185>
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vale karma
that I don't know how to do it
torn jolt
okay can you re-provide the question so that i wont get confused
torn jolt
In this exercise, once I have found "w", how can I obtain the ln(w)^1/2?
its this one right
torn jolt
is this the question you are struggling on?
vale karma
this is what I have done
But now I don't know how to do the ln(w)^1/2
I have to do the root and then, with the four solutions do the ln?
torn jolt
okay give me a second and let me try
glossy valveBOT
Nimajneb
torn jolt
To calculate $\ln\sqrt{w}$ using the principal value of the logarithm, we have:
\begin{align*}
\ln\sqrt{w} &= \ln\left(\sqrt{e^{i\pi/3}}\right)
&= \ln\left(e^{i\pi/6}\right)
&= \frac{i\pi}{6} + 2\pi ni
\end{align*}
glossy valveBOT
Nimajneb
torn jolt
ChatGPT answers
nope
the first part yes
to make sure
the e^itheta part is correct
since the online compiler is not working
second part im struggling
is it that its the angle of difference?
\begin{align*}
\ln(w) &= \ln(re^{i\theta})
&= \ln(r) + i\theta + 2\pi ni \qquad \text{(principal value)}
\end{align*}
where $n\in\mathbb{Z}$ and we have used the fact that the argument of $w$ can be any angle that differs from $\theta$ by a multiple of $2\pi$.
glossy valveBOT
Nimajneb
torn jolt
\begin{align*}
\ln(w)^{1/2} &= e^{\frac{1}{2}\ln(\ln(w))}
&= e^{\frac{1}{2}\ln(r) + \frac{i\theta}{2} + \pi ni}
&= e^{\frac{1}{2}\ln(r)}e^{\frac{i\theta}{2}+\pi ni/2}
&= \sqrt{r}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)
\end{align*}
Im not sure if $e^{i\pi/2} = i$ and $e^{i\pi} = -1$. Maybe therefore, $\ln(w)^{1/2} = \sqrt{r}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)$.
glossy valveBOT
Nimajneb
torn jolt
@vale karma
@torn jolt had to use chatgpt due to spanish
for first part
all other parts are mien
torn jolt
w = ±(cos(30°) + i sin(30°))
w = ± e^(i π/6)
ln(e^z) = z
If you remember this
Well using Euler's formula, we have:
$\cos(30^\circ) + i \sin(30^\circ) = e^{i\cdot30^\circ} = e^{i\pi/6}$
Therefore, i will rather write $w$ as:
$w = \pm(e^{i\pi/6})$
so exp function (exponential)
$w = \pm(\operatorname{cis}(\pi/6))$
glossy valveBOT
Nimajneb
vale karma
ok, I will try finishing it, thanks
torn jolt
By definition, the natural logarithm of a number $x$ is the power to which the base $e$ must be raised to obtain $x$, i.e., $\ln(x) = y$ if and only if $e^y = x$. In this case, $x=e^z$, so we have: $\ln(e^z) = y$ if and only if $e^y = e^z$. But since $e^z$ is the same as $x$, we can write this as: $\ln(e^z) = y$ if and only if $e^y = x$
I will now subsitute it $x=e^z$, we get:
$\ln(e^z) = y$ if and only if $e^y = e^z$
I think tbh $e^z$ is never negative (some rule i dont fucking know) we can take the natural logarithm of both sides of the equation $e^y=e^z$ and get:
$y = \ln(e^z) = z$
Therefore, $\ln(e^z) = z$ for all complex numbers $z$.
glossy valveBOT
Nimajneb
ebon grove
If an object has variable mass giving by the function
$$M(v) = kM_0(v_o - v) \text{if} v < v_o \text{otherwise} 0$$ where v is is the velocity of the object in m/s, k is a constant with value 1 s/m.
$$M_0 = 10kg$$
$$v_o = 100 m/s$$
If the object is dropped from the height of 10m, how long will it take to reach the ground. (Assume g = 10m/s)
glossy valveBOT
roxyit
sharp flame
$\dv{p}{t} = mg$
glossy valveBOT
NEONPerseus
ebon grove
since g doesn't depend on M
sharp flame
Yeah I thought so too but I'm unsure about it
sharp flame
**NEONPerseus**
Nvm I think this is incorrect
keen spruce
2nd derivative of that would be acceleration my guess
keen spruce
can you just use kinematics equations for this
sharp flame
I think there's more to it tbh
terse karma
By solving for $\int_{0}^{T} 10(100-v(t)) 10 dt = 10$ where $v(t)=10t+100$
glossy valveBOT
fäf
terse karma
Most probably wrong
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@ebon grove Has your question been resolved?
harsh quarry
How many possible triplets a,b,c are there such that ab=c, bc=a, ac=b, and all a, b, and c are real numbers?
harsh quarry
How many possible triplets a,b,c are there such that ab=c, bc=a, ac=b, and all a...
a) 3
b) 5
c) 8
d) 9
clear lily
would (x,y,y) be considered a different triplet from (y,x,y)
pliant fractal
Im getting the ans as 4 🫠
grave elm
Im getting the ans as 4 🫠
It must be in form 2+3n
pliant fractal
I just substitued c as a/b
grave elm
I just substitued c as a/b
consider b=0
pliant fractal
Right
harsh quarry
Im getting the ans as 4 🫠
whaa
harsh quarry
I just substitued c as a/b
i got the value of everything as one 🫠
harsh quarry
would (x,y,y) be considered a different triplet from (y,x,y)
i suppose not... they didn't specify anything in the question
clear lily
yeah its sneaky hahah
harsh quarry
yeah its sneaky hahah
yeah lol
wait what
the answer would be 3 right
0,0,0; 1,1,1; -1,-1,1
oh wait
-1,1,-1
1,-1,-1
yeah got 5 i suppose
this is a question from JEE mains
and it has negative marking
So during the exam I think I'd rather just skip the question
harsh quarry
would (x,y,y) be considered a different triplet from (y,x,y)
at this point i think it would be considered different
anyways thanks heavy dude
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clear lily
np
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severe basin
[ g(x,y)=\begin{cases} x^2+e^y & x=y \ y^2+ye^x & x \neq y \end{cases} ]
glossy valveBOT
severe basin
I wanna find the value of the partial derivative of g(x,y) w.r.t y at the point (1,1)
onyx glen
or to put that in less symbols, $g_y(1,1)$
glossy valveBOT
Ann
severe basin
The question actually has the same symbol in it, but I thought people would not know this notation
onyx glen
$\pdv{g}{y}$
glossy valveBOT
Ann
severe basin
I found the partial derivative of the first case and substituted the value in it
That seems wrong
onyx glen
sure is.
that formula only applies on the diagonal line y=x
while g_y(1,1) would need you to find the rate of change in g when going parallel to the y axis
severe basin
I see working out the partial derivative of the second case and substituting values in it, gives the correct answer
But, I don't understand why that works
onyx glen
the region on which the second case applies is everything outside that one diagonal line
so going off of (1,1) along the y axis puts you in that region
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severe basin
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torn jolt
help
torn jolt
i am lost
@cinder cloud
@open harborhelp
<@&286206848099549185>
i’m crying
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unique grove
I thought this graph had an asymptote at y=0, but it crosses the origin? Why is that, and how would I be able to tell?
limber flicker
I think it does have a horizontal asymptote at y=0
unique grove
Yes, so why does it cross the origin?
limber flicker
what part of the definition of a horizontal asymptote states that the graph can't cross the horizontal asymptote?
unique grove
I thought that the behaviour of an asymptote is that the graph will never touch that part of the graph, so if there is an asymptote at y=0, then there will never be a point on the graph where y=0. If that is not the case, then why, and how would I be able to spot that in the future?
limber flicker
this is true for the vertical asymptote
for the horizontal we care about lim x-> inf/-inf f(x) = ?
as the lim x-> inf and -inf (2x)/(x^2-4) = 0 , there is a horizontal asymptote at y=0
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limber flicker
sometimes horizontal asymptotes get called the functions end behaviour, if that also helps picture things
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stuck swan
Hello! I need help with geometric sequence. I need to get the general term aₙ if a₁ - a₃ + a₅ = 651 and a₁ + a₇ = -325. I tried to solve it, and I will show my steps below, but I got stuck.
a₁ - (a₁ x q^2)+(a₁ x q^4) = 651
a₁ + a₁ x q^6 = -325
I could extract a1 from first formula but idk what to do after
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Hello! I need help with geometric sequence. I need to get the general term aₙ if...
@stuck swan Has your question been resolved?
rain plinth
Hello! I need help with geometric sequence. I need to get the general term aₙ if...
divide 1 equation by the other
also you can find q^2 instead of q because all the powers are even
stuck swan
divide 1 equation by the other
also you can find q^2 instead of q because all th...
I'll try that, thanks
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storm goblet
Determine the center and radius of the circle if the equation of the circle is x^2+y^2-10y=0
storm goblet
I know the center is 0, 0
But how do i get the radius
@viral jasper what is the center
viral jasper
Complete the square for y^2-10y
storm goblet
How do i do either of those things
torn jolt
Let's start simple
Typically when you have the equation for a circle the second term is (y-b)^2
What do you get if you expand (y-b)^2?
storm goblet
Nothing it's just y^2
storm goblet
y^2-2yb+b^2
torn jolt
If you follow along I'll show you how this helps you solve your question
torn jolt
y^2-2yb+b^2
We'll write 2by instead for the sake of simplicity. In your equation, you have y^2-10y. So, to match this, what b should you pick?
(ignore the +b^2 for now)
ember shadow
you'd have a lot fewer questions if you checked out a course or book, like was suggested yesterday multiple times: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:more-on-completing-square/a/completing-the-square-review
storm goblet
x^2
torn jolt
x^2? x shouldn't even come into play here
storm goblet
10y
storm goblet
5
torn jolt
Indeed
rain plinth
now complete the square by adding something to both sides
severe basin
Considering the general form of equation of a circle makes it easier, no?
rain plinth
you'd have a lot fewer questions if you checked out a course or book, like was s...
true
torn jolt
(y-5)^2 would give you y^2-10y+25. That +25 which is the +b^2 we were ignoring this whole time is going to be a problem... We should get rid of it, do you know how?
storm goblet
x=-y+5?
torn jolt
I still don't know what x is doing here
Plus we aren't looking to solve for x here
x is just a variable, not an unknown
storm goblet
-25?
torn jolt
Oh actually I guess you can have it equal to something dependent on y but whatever
severe basin
This can help you answer the question without completing the square.
torn jolt
(y-5)^2 would give you y^2-10x+25, hence (y-5)^2 would give you y^2-10x+25-25=y^2-10x
Does everything so far make sense?
torn jolt
We do, but that only comes into play once we have the full equation in standard form
To recap, your equation had y^2-10y, which is not in the standard form for the equation of a circle. So, we used the fact that (y-b)^2=y^2-2by+b^2 and realised that b must be 5, which gives us (y-5)^2=y^2-10y+25. To get rid of that +25 and get back the equation we had at the beginning, we removed 25, which gave us (y-5)^2-25. The reason we did all this was to get something closer to the standard form for the equation of a circle, which is what we now have
Now your original equation becomes x^2+(y-5)^2-25=0
There is one last step to get this into standard form, do you see it?
storm goblet
we need to turn 25 into 2 * x * y?
storm goblet
I do not see it
torn jolt
What is the standard equation for a circle?
storm goblet
(x-a)^2+(y-b)^2=r^2
torn jolt
Right, and do you see how x^2+(y-5)^2-25=0 differs from the standard form?
torn jolt
If you did this it wouldn't be the same equation
storm goblet
well i still don't get the center
torn jolt
well i still don't get the center
Okay?
...So what do you get if you move it to the right?
storm goblet
or the radius
torn jolt
Just
just
i-... What does the equation become of you do what you just said you should do?
storm goblet
x+y-5=5
torn jolt
the what
storm goblet
x^2+(y-5)^2=25
torn jolt
You must've eaten the squares
Ah there it is
Oh I see
You thought you could take the square root of both sides and the squares would all go away
torn jolt
Unfortunately sqrt(x^2+(y-5)^2) is not the same thing as sqrt(x^2)+sqrt((y-5)^2)
severe basin
a²+b²=c² does not imply a+b=c
storm goblet
okay what do i do next
torn jolt
x^2+(y-5)^2=25
Well there is one last detail that makes this differ from the standard equation
storm goblet
-5
torn jolt
No, that's part of the standard form
storm goblet
25
storm goblet
5^2?
storm goblet
What's next
storm goblet
How do i read off the values
storm goblet
5
torn jolt
What is b?
storm goblet
5
torn jolt
And, lastly, what is a?
Hmmm I see this one is not as obvious
You'll feel stupid once you find out (I don't mean this as an insult)
For what value of a is (x-a)^2=x^2 for all x?
storm goblet
0?
torn jolt
Yep
storm goblet
r 5 and c 0,5
torn jolt
Mhm
storm goblet
Can you recap
torn jolt
Good idea
So we started with x^2+y^2-10y=0
Our goal was to transform this equation into the standard form, which is (x-a)^2+(y-b)^2=0
First of all we noted that (y-b)^2 expands as y^2-2by+b^2
And hence concluded that we must match 2by with 10y, which is only possible if b=5
Also last time I mentionned this in the end but we can clearly see that a must be 0 here
(y-5)^2 expands as y^2-10y+25. We got y^2-10y as we wanted, but as a side effect an unwanted +25 popped up, so we canceled it out with a -25
This is how we figured out that y^2-10y=(y-5)^2-25
All put together, our equation is now x^2+(y-5)^2-25=0
To finish this off, we moved that 25 to the other side and wrote it as 5^2
Our final equation is x^2+(y-5)^2=5^2, which is in standard form, and hence we can just read off the values now
All you have to remember is that x^2=(x-0)^2
By the way, if you ever get something like (y+5)^2 where we have a + instead of a -, don't be fooled and read this off like a 5. It's secretly (y-(-5))^2, so the value you should read off is -5, not 5
@storm goblet Does all of that make sense?
torn jolt
@storm goblet Don't forget to .close the channel
storm goblet
.close
alpine hinge
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alpine hinge
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rain plinth
how did they go from that to that
sin(A)cos(B)/cos(A)cos(B) = tan(A)
cos(A)sin(B)/cos(A)sin(B) = tan(B)
and same for bottom
distribute division
alpine hinge
distribute division
wdym
like first two in a fraction
plus the second two in a fraction?
rain plinth
you divided the numerator and denominator by something
you can distribute division over addition
divide each term by the divisor
rain plinth
you can distribute division over addition
(just like multiplication)
alpine hinge
you divided the numerator and denominator by something
how did they know to do that
divide by that
alpine hinge
what 😭
rain plinth
if you compare sin(A)cos(B) to tan(A)
you will see that the ratio is cos(A)cos(B)
and if you compare cos(A)sin(B) to tan(B) the ratio is the same
so you might start thinking
hmmm if you multiply the top by this thing it works maybe it works on the bottom
and so it does
and then you know it worked
alpine hinge
if you compare sin(A)cos(B) to tan(A)
you will see that the ratio is cos(A)cos(B...
i can also just
multiply
by that as well
right
like division it’s te sam etching
?
alpine hinge
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errant portal
Can someone check this
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Can someone check this
errant portal
$81.60 Agri
$138.23 manufacturing
$199.89 services
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pulsar gale
@errant portalwhy do you think pinging will still help
pulsar gale
$81.60 Agri
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we arent really here to check your work
just submit and see what happens
torn jolt
we arent really here to check your work
ic
errant portal
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bronze magnet
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bronze magnet
i found the lower estimate by moving by increments of 2 to the right
i need to find the upper estimate using 5 rectangles as well
i am given the 10 exact coordinates for the 1 - 10 points
how would i find my upper estimate using 5 rectangles?
would i move to the left?
Yes
ty
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warm pier
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warm pier
Could anyone help me with these problems?
cedar coral
we have the rule $(ab)^c=a^c\cdot b^c$
glossy valveBOT
ThisGuy
cedar coral
we know, for example the square root $\sqrt(x)=x^{\frac{1}{2}}$
glossy valveBOT
ThisGuy
cedar coral
likewise, the fifth root is ^1/5
cedar coral
with these rules, seperate your roots
torn jolt
does |A| = A^-1 (just different ways of expressing it)
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does |A| = A^-1 (just different ways of expressing it)
torn jolt
inverse matrix
torn jolt
oh so |A| is the determinant, A^-1 is inverse
cedar coral
yeah
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torn jolt
shit
cedar coral
i was just saying, |A|=A^-1 doesnt make sense for matricies
cause |A| is scalar, and A^-1 is a matrix
torn jolt
yeah i read the notes wrong i think
torn jolt
cause |A| is scalar, and A^-1 is a matrix
makes sense
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fallow jacinth
hmm Im still getting something wrong
ok I got it
What about the other one
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fallow jacinth
Can I get some insight on why my answer is wrong? Did I do the integral wrong?
brittle steeple
again, I don't see any -11.58x term in f(x)
did you properly integrate f'(x)?
or did you integrate the version of f'(x) where C is 0
fallow jacinth
@brittle steeple right. I took my answer for the first question and integrated that. I then checked the integration with a integral calc and it is right but something is still wrong
fallow jacinth
its ok this website doesnt need to simplify
brittle steeple
f(1) is supposed to be 8, right
and currently it is 2.75
so what do we need to add to it to make it 8
fallow jacinth
it simplifies by itself
fallow jacinth
f(1) is supposed to be 8, right
-8
brittle steeple
ah ok
so it's supposed to be -8
and currently it is 2.75, before we add a constant?
light sonnet
it simplifies by itself
What happened to that -11.58x?
fallow jacinth
light sonnet
I'm referring to in your answer, on that webpage. Where is the -11.58x?
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torn jolt
Find n as a positive integer such that n^3 - 5n + 18 is a power of 2
obtuse epoch
can somebody help me this I dont know how to explain
hot herald
@obtuse epoch this channel is occupied. please read #rules #❓how-to-get-help and claim your own channel
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elfin stream
Find n as a positive integer such that n^3 - 5n + 18 is a power of 2
Well u have two options
A) use the ugly cubic formula
B) factorise and then set it equal to 2^x
torn jolt
Well u have two options
A) use the ugly cubic formula
B) factorise and then set...
how do i factorize it?
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quaint pollen
can anyone help ?
restive hawk
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restive hawk
hello,, dont really have time to actually solve this but could someone let me know if ive set up part b correctly?
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torn jolt
i cant do related rates
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i cant do related rates
torn jolt
here is the question
keen spruce
GIVEN
FIND
EQUATION
torn jolt
a wompan who is 156 cm tall walks directly away from a street light that is 4.32 m above teh ground. when she reaches a point on the sidewalk that is 12.6 m away from the base opf the lightpost, her shadow is lengthening at a rate opf .85m/s how quickly is the woman walking
sry for my spelling mistakes
trying to type fast
onyx glen
156 meters tall?
onyx glen
trying to type fast
what's the rush
torn jolt
156 meters tall? <:hmmCat:831508801646755850>
cm*
torn jolt
what's the rush
this is overdue for me
its pising me off i cant learn this
i want to learn it fast so i can get through a few assignments
onyx glen
"learning" and "fast" are usually antithetical.
torn jolt
yes
onyx glen
and if you are pissed, you will have a harder time learning.
torn jolt
but it needs to be done
torn jolt
and if you are pissed, you will have a harder time learning.
yes
some people told me to step away from it after 10 mins
that made me even more mad
anyways
what i did
is create a triangle
used similar triangles
i found the distance at that moment from the woman to the end of the shadow to be 7.12 m
i have no idea what to do now
its gonna be some pythag then take deriv
but i dont know what to compare
light sonnet
some people told me to step away from it after 10 mins
You understand that most people who said this, they meant it as a chance to give yourself a break, so you can feel more refreshed and start over with a clearer mindset, right?
torn jolt
You understand that most people who said this, they meant it as a chance to give...
yes, but been trying for days
its a new day for me
i need to try again
onyx glen
got a diagram?
torn jolt
yes
torn jolt
sec i need to turn on the wifi
also one thing to note
i dont know if i even labelled things correctly
which is likely why its wrong
torn jolt
ok show it here
^
onyx glen
sorry my internet shit itself
torn jolt
np
onyx glen
the length of the shadow is w not S
torn jolt
the length of the shadow is w not S
i dont understand that though
since dw/dt is the speed of the woman
onyx glen
your notation is incorrect
the shadow is on the ground
its length is the length of that horizontal segment you labeled w
torn jolt
ok i see that
but where is the woman
like the place to calc the speed
i dont get how it would be in the air
or is that just not it
onyx glen
i'm on a train so can't rly draw the correct diagram unfortunately.
torn jolt
i'm on a train so can't rly draw the correct diagram unfortunately.
could you please describe it?
onyx glen
sure.
draw a horizontal line to represent the ground. on it, mark in order from left to right two points P and Q; mark the length of PQ as 12.6.
from P, draw a vertical line PL upward to represent the lantern. from Q, draw a vertical line QW upward to represent the woman; mark PL and QW with their lengths.
extend LW to meet PQ at R.
the shadow will be QR.
torn jolt
sure.
draw a horizontal line to represent the ground. on it, mark in order from...
alright it looks the way you described it now
ill send it
onyx glen
the shadow should be proportional to the distance between the lantern and the woman
iirc
torn jolt
torn jolt
the shadow should be proportional to the distance between the lantern and the wo...
yes the ratio works
but now i dont know what to solve for
well thats a lie
i know what to solve for
but i dont know the equation
How would i find it
?
onyx glen
so you found QR = 7.12
so then the shadow is (7.12/12.6) times the distance from the lantern
so their rates of change are proportional too
with the same ratio
torn jolt
so then the shadow is (7.12/12.6) times the distance from the lantern
how will this help though?
onyx glen
literally you know the relationship between their rates of change
torn jolt
so then .85 /that ratio?
why would i even be dividing tho
it gives 1.5m/s
which is a correct answer
but im not sure why im dividing
grave elm
but im not sure why im dividing
try to write the "the ratio" as d something/d something
torn jolt
try to write the "the ratio" as d something/d something
Im not sure
because im just comparing base lengths of the triangle
grave elm
do you know what that ratio represents? Can you explain it in natural language?
torn jolt
$d(b_1)frac/d(b_2)?$
glossy valveBOT
Caleb.
torn jolt
do you know what that ratio represents? Can you explain it in natural language?
the length of the lightpost to the woman compared to the length of the woman to end of the shadow
12.6/7.12 * ds/dt = dw/dt
grave elm
well so the equation is all you need, ds/dt is 0.85
grave elm
as you wrote here earlier
grave elm
the length of the lightpost to the woman compared to the length of the woman to ...
and since length of shadow is proportional to distance of the woman from lamp, 12.6/7.12 represents also dw/ds. So dw/ds*ds/dt=dw/dt, since ds cancels out. So the equation you wrote is true
torn jolt
and since length of shadow is proportional to distance of the woman from lamp, 1...
wow ok
i didnt realize that
very cool
thank you @grave elm @onyx glen
grizzled ridge
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grizzled ridge
good day people can anyone confirm if my answers are correct>
?
For Zu=8
For Zuv = -1
spice orchid
Easier and quicker if you send you own work and then people can check that
grizzled ridge
grizzled ridge
Easier and quicker if you send you own work and then people can check that
sorry for that, here sir
grizzled ridge
disregard the "u" for the Zuv, second line, 3rd and forth term
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devout inlet
$\pdv{z}{u} = v f_x + 2u f_y$
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stabulo
devout inlet
is what I get but you have something different.
By equation (3) adjusting for the different letters you should get the result I posted above.
Interestingly though the correct procedure produces the same answer.
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viral stratus
q3
viral stratus
how do u do it
this is the answer
this is what i did
idk what i did wrong
someone please help, is it the bounds
copper lava
Give me around 2-3 years tops. I'll most likely understand it by then. Best of luck.
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frank vault
hold on im doing it
viral stratus
oh ok, sorry tysm
viral stratus
why
frank vault
hold on
because ur integrating from angle to angle
the first part is integrating r=cos(theta)
nvm not cos theta
i did it backwards
there
frank vault
in some cases for integrating polar regions you have to add
and then u have to have different bounds
frank vault
good
frank vault
gotchu
tacit stone
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tacit stone
Hi, I need help with 4b
Specifically, it is not clear how I would construct the smallest unique topology that includes all of T_alpha
And if i do construct it, how I would show its uniqueness
stiff musk
try taking the intersection of all topologies on X that contain all of the T_alpha
you know that intersection is not vacuous because the discrete topology on X is one such topology
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tacit stone
you know that intersection is not vacuous because the discrete topology on X is ...
haha yeah, i forgot to closee this channel bcs i alr got the answer
Thanks!
queen urchin
Bit lost on how to start this
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Bit lost on how to start this
wise wyvern
You can expand this at x = 1.
you'd get tan^(-1)(1) = expansion at 1.
And the lhs would be pi/4 and depending on how good you want the estimation, write as many terms.
queen urchin
You can expand this at x = 1.
you'd get tan^(-1)(1) = expansion at 1.
And the ...
What exactly do you mean? I got what the expansion overall is but not sure how to relate that to pi or 1/5
wise wyvern
tan^-1(1) = pi/4
And the rhs would be the expansion at x = 1
So if you can approximate pi that way.
queen urchin
So if you can approximate pi that way.
Ah interesting, I think I see what you mean. But is there any way to specifically relate it to 1/5 or do I just have to put enough terms so that we can just see that the error is lower?
wise wyvern
Personally, I would say the latter. Unless of course someone else has another way.
queen urchin
no lagrange error bound or anything?
wise wyvern
This was the best I had.
queen urchin
Ah ok. Thank you
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queen urchin
If anyone else has a thought go ahead
queen urchin
What series test do I need to use to show convergence of 1/(n^ln(n))?
queen urchin
Ratio test?
strong crystal
Personally i think the integral test
Actually no ratio test is fine
Itll work and is easier than integral test
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queen urchin
Itll work and is easier than integral test
Ah ok. Will try
craggy sandal
I need to prove the Euler totient function is multiplicative using Euler's product formula
craggy sandal
So $\phi(nm) = nm \prod_{p | nm} (1 - \frac{1}{p})$
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LeftySam
craggy sandal
but idk where to go from here really
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fast peak
so you know that n and m are coprime. what does that tell you about p|nm
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vague cradle
Is this solved correctly
The task is to draw Set S
Is this how I do conj(z) with Arg?
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amber radish
So your question is this right?
Just a confirmation
i.e. find the geometric interpretation of the set S on the complex (aka Argand or Gauss) plane
amber radish
So your question is this right?
this is meant to say region instead of locus
shell dawn
quick question, if I have a power tower, as in x^x^x... and the whole thing square root. If i write the root as power, ^1/2 is placed at the base of the tower or top?
shell dawn
$\sqrt{x^{x^{x}}}$
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LILY
night folio
quick question, if I have a power tower, as in x^x^x... and the whole thing squa...
You multiply the exponent of the base by 1/2
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LILY
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
night folio
Yes
normal thicket
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white stream
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onyx glen
oh god not one of those
white stream
I tried to take the difference bet 2 terms but cant quite find a result
onyx glen
you cannot hope to arrive at an answer here except by guessing.
white stream
I sent it to chatGPT but it found out -25, -17
white stream
surely
umbral kestrel
i made it calculate my final grade of all the subjects and it failed to sum the terms haha
it is very very bad.
lean rock
you can make it more accurate: https://github.com/openai/openai-cookbook/blob/main/techniques_to_improve_reliability.md
onyx glen
you can make it more accurate: https://github.com/openai/openai-cookbook/blob/ma...
bandaid measures, surely.
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lean rock
let's solve this
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