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#help-23
worn relic
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void crag
i have a question regarding the 7th rule it says here that L has to be grater than 0 i know that it cant be negative since even root function cant be negative but why cant it be 0?
safe radishBOT
i have a question regarding the 7th rule it says here that L has to be grater th...
void crag
was it because when x approach c from left hand rule its negative?
safe radishBOT
@void crag Has your question been resolved?
junior smelt
was it because when x approach c from left hand rule its negative?
Yes, you won't have the left limit (because from the left, your function (f(x))^{1/n} won't be defined) 
slate notch
safe radishBOT
slate notch
how to do b?
lean otter
V(x) = Volume, correct?
slate notch
yurrr
silent imp
set v=1000 and solve for x
slate notch
I did
lean otter
You can substitute 1000 for V(x)
slate notch
problem is idk how to factor cubic functions
lean otter
And foil out
slate notch
textbook never taught
lean otter
There is a formula for it
lean otter
its an extending question
ok
slate notch
school wouldnt make you do somthing they didnt teach
u wouldnt be allowed to search for a formula in a test
lean otter
I don’t Even think you need to uhm multiply the foiling by x
silent imp
you can also differntiate
slate notch
luckily, I have a calculator that can solve x for me
lean otter
u wouldnt be allowed to search for a formula in a test
Ah right
slate notch
but if I need to show my work that will be an issue
idk if the school will even give htis question but well see
wont seee this for another year or so
slate notch
you can also differntiate
how
slate notch
kk
lean otter
600 -100x+4x2, you can arrange that into a quadratic equation, and I don’t think you need to multiply the entire thing by x
slate notch
silent imp
can you show me the final cubic formula you found
slate notch
this textbook last question of 3.3
slate notch
can you show me the final cubic formula you found
4x^3-100x^2+600x
silent imp
and then you have -1000 right
lean otter
Yes I was going to say use that
slate notch
quadratic equation is always automatically set to equal zero
you cannot just set it to another number lol
lean otter
quadratic equation is always automatically set to equal zero
So u move the 1000 to the other side to set it to 0
slate notch
u forgor to multiply it by x
lean otter
wait
lean otter
u forgor to multiply it by x
Yea mb
silent imp
So u move the 1000 to the other side to set it to 0
but you cant use the quadratic then
lean otter
but you cant use the quadratic then
Yea I realized my mistake
I think you’ll have to factor it
slate notch
school wont give this so I most likely wont have to wory lol
lean otter
school wont give this so I most likely wont have to wory lol
okay that’s good
do attempt to factor it though
slate notch
idk tho some teachers are just mean and give stuff a year ahead they didnt teach
silent imp
you can factor out x-5 from the final cubic
slate notch
idk how that qualifies as a thinking question though
uncut star
what about using the pq formula?
slate notch
as its just straight up somthing they didnt teach
slate notch
what about using the pq formula?
didnt learn yet
lean otter
what about using the pq formula?
I don’t think this person had went through that yet
oh ya
slate notch
you can factor out x-5 from the final cubic
I saw that solutuion
but idk how
I wanna understand it lol
silent imp
but idk how
That doesnt havee how, you just see and know when you have a cubic, idk how to explain it but it doesnt have any reason, you see it and you guess
silent imp
That doesnt havee how, you just see and know when you have a cubic, idk how to e...
when you encounter more and more cubics, your guess becomes more accurate
and there were also methods involving calculus, but for cubics like this factoring will be the easiest
safe radishBOT
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manic path
could someone tell me in word what this means
safe radishBOT
could someone tell me in word what this means
manic path
eternal carbon
s is the set of three-dimensional column vectors such that [that equation] is true
peak estuary
the equation says that x is orthogonal to the gradient
manic path
yea but I didn't give the whole context they just want me to prove it's in the subspace
I know how to prove it but math syntax sucks
eternal carbon
the equation says that x is orthogonal to the gradient
oops
manic path
wait just to understand quicker
we are given F(x,y,z) should I partial diff those and sub it into the original equation to prove it's a subspace or can I just do it with that
alr nvm
.close
steady jolt
safe radishBOT
steady jolt
bit lost on what to do
rough storm
,rccw
flat frigateBOT
lean otter
area of the triangle - area of the semicircle
eternal carbon
bit lost on what to do
you can do this by performing [area of the triangle] minus [area of the circular sectors]
steady jolt
oh is that it?
rough storm
mhm
lean otter
oh is that it?
yup
uncut star
ballin
rough storm
yes, given theta is in degrees
steady jolt
it has 2 radius'es
rough storm
it has 2 radius'es
?
steady jolt
there are two lines on either end where the radius is
lean otter
it has 2 radius'es
no
steady jolt
two sets
lean otter
yes but the radius is same for all of them
steady jolt
oh
rough storm
those two lines just mean those line segments are equal in length
lean otter
so just add up all of their angles and put in the formula
visual timber
hello, im new here; i just started to learn math, and i was wondering if this is the right place to ask for advice? i've came across khan academy, and im doing "pré-algebra", (i found it by searching in what order i should learn the subjects) but i dont feel like im learning much from it, are there other better or just different ways of learning? should i go through textbooks, try to solve the problems and search what i dont understand, watch lectures on yt? also if im wrong on something pls lmk; ;
safe radishBOT
hello, im new here; i just started to learn math, and i was wondering if this is...
eternal carbon
can you provide a few more details on how khan academy doesn't seem to be working for you?
visual timber
can you provide a few more details on how khan academy doesn't seem to be workin...
ty for replying, yes. i can't apply the information from the lessons on the questions; sometimes i will get the answer right, but wont be able to solve it again (aka not truly understand it), i try to search for different tutorials on the same subject if the video is confusing, but its still really difficult to grasp how to interpret the questions. also the lecture > quizz method seem pretty demotivating, but maybe thats a me issue? im not sure if theres a different way to do it; ;
safe radishBOT
@visual timber Has your question been resolved?
pliant patio
Don't know where to start here...
safe radishBOT
Don't know where to start here...
brave wolf
Tryu-substitution
try turning everything that's inside of f into u
(in the second integral)
pliant patio
thanks but erm what
wind laurel
when you evaluate the first integral
you will get F(8) - F(4)
F(x) being the antiderivative of f(x)
pliant patio
ye
wind laurel
so F(8) - F(4) = 5
pliant patio
ye
wind laurel
now for the second integral, evaluating it you will get
F(2(2+2)) - F(2(0+2))
or
F(8) - F(4)
oh
wait 5 is not an answer
brave wolf
wait 5 is not an answer
yes
brave wolf
now for the second integral, evaluating it you will get
F(2(2+2)) - F(2(0+2))
This isn't true, because F is not necessarily antiderivative of f(2(x+2))
brave wolf
going in circles
Do you know u-substitution by chance?
wind laurel
ah
pliant patio
Do you know u-substitution by chance?
No
brave wolf
hmm
then we can think about function trasnformations maybe
imagine a graph of f(x)
What would graph of f(2(x+2)) look like?
What would be the steps you would make to transform graph of f(x) into graph of f(2(x+2))?
pliant patio
double the x values translated 2 units left?
brave wolf
double the x values translated 2 units left?
what do you mean by double the x values?
do you know what would graph of f(2x) do to graph of f(x)?
pliant patio
double the y values
brave wolf
not quite
have you been taught function trasnformations?
and are you 100% sure you havent been taught u substitution?
brave wolf
and are you 100% sure you havent been taught u substitution?
that's the method to reverse chain rule basically
pliant patio
yea but I'm a professional crammer so I don't recall that stuff well
pliant patio
have you been taught function trasnformations?
in response to this
brave wolf
Alright
this graph shows f(x) as black line and f(2x) as red line
can you see what f(2x) does to the graph of f(x)?
pliant patio
right
brave wolf
what do you think happens to the area when you do this?
pliant patio
halves?
brave wolf
yes
pliant patio
right
brave wolf
if you also squish the points between which the area is calculated
and the (x+2) part does what to the graph?
pliant patio
translates2 units left
pliant patio
nothing?
pliant patio
yep
brave wolf
what can we say about area under f(2x) between certain 2 points? (And between what points?)
brave wolf
keep in mind that it squishes the graph
pliant patio
so it will be the same as the area between x=0 and x=2
pliant patio
yea
brave wolf
this is some f(x) with the area marked on it
now we squish it by a factor of 2
and we squish it towards 0
so the point which is now at 2 will end up at 1
the point which is at 4 will end up at 2
etc
pliant patio
right
brave wolf
so where will the current bounds 4 and 8 end up?
pliant patio
2 and 4
pliant patio
half
pliant patio
x=0 and x=2
pliant patio
not change?
pliant patio
so its 5/2
brave wolf
correct, i gtg now, sorry
tender dawn
hi
tender dawn
how do i solve the integral
∫ x^4 * (1-x)^3 dx
yes i can multiply everything. but like without doing that
thin bridge
by parts it a few times
tender dawn
what would be step 1?
thin bridge
choose which component you want to integrate / differentiate
thin bridge
choose based on whats easier to integrate and differentiate
tender dawn
none of the methods worked
U = (1-x)^3 gives me nothing
and U = x^4 gives me nothing
so i am asking, what should i pick then
thin bridge
wdym gave you nothing
you'll get something in both cases
and if you don't want to expand, by parts again
this is one way to approach the problem if you really don't want to expand everything out
thin bridge
show what you have
tender dawn
du = -3(1-x)^2 dx
thin bridge
it looks like you're approaching this with u-sub / chain rule
which isn't what i said
tender dawn
@thin bridge it didn't do anything, now i just have ^2 instead of ^3 but everything looks way more messy...
thin bridge
so yeh clean it up a bit and apply by parts again to the second integral
you'll then get ^1, repeat again and again
till you get a simple power function
tender dawn
but what if the intregral started with (1-x)^20, are u telling me i would have to repeat this 20 times??
thin bridge
a lone (1-x)^20 or something multiplied to it?
tender dawn
x^4 * (1-x)^20
thin bridge
well you wouldn't need to repeat 20 times for that
tender dawn
okay
thin bridge
how many times you'd do by parts for something like this depends on how many times you'd differentiate your component to get something like 1
so in that case you can choose to differentiate the x^4 component,
since (1-x)^20 is relatively easy to integrate and won't cause major problems
tender dawn
okay thanks for help
thin bridge
tabular format of ibp also keeps things more organised
tender dawn
.close
cursive scaffold
Could i ask where I went wrong?
Was trying to simplify a Boolean Algebra~
I can't seem to figure out where the mistake is, It's past the distribution law at the end but I am having a hard time understanding
safe radishBOT
Could i ask where I went wrong?
Was trying to simplify a Boolean Algebra~
I can'...
cursive scaffold
Distributive law should be correct
So should the idempotent law afterwards
Then the second last step?
But then what did I do wrong 🤔
Can u not take it out common like that 👀
short topaz
Could i ask where I went wrong?
Was trying to simplify a Boolean Algebra~
I can'...
i think the final answer is just A and B or B and C
or alternatively, B and (A or C)
i think you went wrong in the line after (A and B) or (B and C)
cursive scaffold
i think the final answer is just A and B or B and C
Hmm hmm
cursive scaffold
i think you went wrong in the line after (A and B) or (B and C)
I think it's the second last line~
Where I took the A out common?
That isn't possible, but then it feels difficult to wrap head around these tbh
cursive scaffold
Yep, did end up that way
But it still feels a little confusing though
How could I actually go about approaching this?
I feel like I wouldn't really catch where I might have gone wrong
short topaz
i think the easiest way to finish the problem is to spot that
(A and B) or (B and C) is the same as
(B and A) or (B and C)
so it factorises as B and (A or C)
right i see what you've done now
cursive scaffold
How do u do the third line here though
B and A and B or C
short topaz
i mean the best way is just to think about it
cursive scaffold
👀
short topaz
however, if they require you to have like "law of X" for each line
then once you've thought about what the answer should be
cursive scaffold
How did u get the B out from here 👀
short topaz
you can usually find the steps required to get to the final answer
cursive scaffold
i mean the best way is just to think about it
Think about it?
What the simplified form would be?
Nvm
short topaz
nw
i think like normally, with these problems
if you don't need like some thorough justification for each step (i.e. ur simplifying a set), then you can normally think about it logically
rather than trying to remember n different rules
but if you need to give like what axiom ur using for each line
cursive scaffold
I believe we will have to
short topaz
once u've thought about how it should simplify, you can then find the steps to do so using 1 axiom at a time
short topaz
once u've thought about how it should simplify, you can then find the steps to d...
normally you can like add something redudant to help you simplify an expression
cursive scaffold
Hmm
short topaz
reason why i mention this is i think if u realised that the answer could be written as B and (A or C) earlier
short topaz
(A and B) or (B and C) is the same as
you might have been able to spot this earlier
which would've been easier than trying to expand that
cursive scaffold
Here do u B and B to get rid of the additional B inside the bracket 👀
short topaz
yeah basically
cursive scaffold
you might have been able to spot this earlier
Hmm 👀
I think where it actually went wrong is this thing though~
The others while not ideal steps, probably aren't wrong right?
short topaz
right i see what you've done now
oh yeah i should also say where u went wrong
all ur steps were fine until
short topaz
(A or B) and (A or C) factorises as A or (B and C)
this
you wrote A and (B and C) instead of A or (B and C)
cursive scaffold
short topaz
as in (A or B) and (A or C) and B = (A or (B and C)) and B
but tbh, you could've also done
cursive scaffold
you wrote A and (B and C) instead of A or (B and C)
Ah
short topaz
B and (A or B) and (A or C) = (B and (A or B)) and (A or C)
B and A and (A or C)
etc.
cursive scaffold
you wrote A and (B and C) instead of A or (B and C)
Thank you though
short topaz
nw
cursive scaffold
How should I approach thinking about the simplified form though?
I am having a hard time imagining these lines of text explaining the issue, How do u visualise these in a fluent manner 🤔
Is it just getting used to the notations and growing familiar over time?
safe radishBOT
Available help channel!
signal forge
Hello
Is there a function that when multiplied by its inverse, it would yield one?
$f(x) * f^{-1}(x) = 1$
safe radishBOT
Hello
Is there a function that when multiplied by its inverse, it would yield on...
flat frigateBOT
Stella
spice furnace
Hello
Is there a function that when multiplied by its inverse, it would yield on...
no what you are dealing with is functional inverse not a multiplicative inverse
stoic dune
I'd experiment with functions of the form (ax+b)/(cx+d)
signal forge
no what you are dealing with is functional inverse not a multiplicative inverse
what i'm asking is, is this possible?
gusty trench
if f(x) = x^n, then its inverse would be f^-1(x) = x^(1/n). then you'd want x^n * x^(1/n) = 1 for all x, which means that n + 1/n = 0, or n^2 = -1
it looks like f(x) = x^i and f(x) = x^(-i) are two such functions satisfying this
specifically because of the property that the reciprocal of i is exactly its opposite, causing the addition of the powers from the multiplication to produce 0
signal forge
so it has no solutions with real numbers?
gusty trench
i haven't been able to think of one, there might be
stoic dune
Let f={(1, 1)}
One could argue this is a solution with real numbers. In case you're not clear on the notation: this is the function that only accepts 1, and maps it to 1.
But I get if you want one with continuity somewhere
signal forge
But I get if you want one with continuity somewhere
more like one with an algebraic equation
signal forge
https://www.quora.com/How-could-f-1-x-frac-1-f-x
yeah this answers it
thanks!
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safe radishBOT
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Available help channel!
rain dagger
why <=
n >= 1
safe radishBOT
why <=
n >= 1
vale oriole
im confused about the question, are u asking why they use <= if n >=1?
rain dagger
im confused about the question, are u asking why they use <= if n >=1?
why the inequality holds true
vale oriole
for n>=1?
faint seal
$\log(1+x)\le x\forall x>0$
flat frigateBOT
kheerii
faint seal
if you want a proof of this, you can create a function f(x) = log(1+x) - x and check the derivatives of f to see that the function always decreases
and since f(0)=0, f(x) < 0 for all positive x, meaning that log(1+x) < x for all positive x
rain dagger
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still lily
im not sure on how to get the end points and the length of the LR, and if my points are correct on the graph,, and im not sure if everything i answered is right 🥹
safe radishBOT
im not sure on how to get the end points and the length of the LR, and if my poi...
@still lily Has your question been resolved?
orchid ledge
safe radishBOT
orchid ledge
i think i know the answer but i want to know how to solve it
orchid ledge
ok thanks
brave wolf
lmk if you got stuck somewhere
safe radishBOT
@orchid ledge Has your question been resolved?
stone sedge
Hello I was just wondering, what’s the difference between a sine and a cos graph??
safe radishBOT
Hello I was just wondering, what’s the difference between a sine and a cos graph...
timber karma
cos is 1 at x = 0 while sin is 0 at x =0
soft matrix
Hello I was just wondering, what’s the difference between a sine and a cos graph...
Both are same waves but with a phase difference of pi/2
stone sedge
So cos starts at 1 on the y axis if I graph it and Sine starts at 0?
stone sedge
ohhh alright
safe radishBOT
@stone sedge Has your question been resolved?
tropic gale
Not sure if I should ask this in another discord for physics help, but this is a question I came up with myself that I would like to know the answer for because I got absolutely no idea:
safe radishBOT
Not sure if I should ask this in another discord for physics help, but this is a...
tropic gale
I have two identical cars. In each car, 4 weights of 50kg each are placed. One time in the center and one time on the edges. The center of mass remains the same. However, will the turn-rate of these cars be different? How will weight-shifting during braking into a corner affect this if at all?
hardy lion
are you familiar with moment of inertia?
tropic gale
are you familiar with moment of inertia?
Yes. The turn on a chair with weight in your hand thing, right? But cars are not turning on a fixed point so I am confused if this would apply here
hardy lion
ok, so if the car is turning around a corner, the car on the right might turn slightly slower because moment of inertia is based on r^2
and if the car is spinning on ice, the car will turn significantly slower because the center point is the center of mass
you may get a more informative response in discord.gg/physics
tropic gale
you may get a more informative response in discord.gg/physics
Will join thanks. But on here, when you brake hard into a corner, the weight of the car shifts forward, changing the center of mass. I was wondering if this would excarbate the slower turning on the right car or be unrelated
hardy lion
im not exactly sure what you mean
tropic gale
im not exactly sure what you mean
During slowing down for turning, the front-suspension of a car gets loaded, shifting the balance of a car to the front. During turning in, the balance will shift to the side, to the outside of the corner (away from the turn). I am asking for this being more pronounced if the weight distribution is on the outside of the car
still surge
what do you mean front suspension of a car gets loaded?
the weight distribution is still the same and does not change the center of mass
hardy lion
i would think the weight being more evenly distributed around the car would reduce the load on the front
but im not sure
still surge
https://en.wikipedia.org/wiki/Weight_transfer
erm yeah I'm not too polished in this but this paragraph in the wiki might help Lowering the CoM towards the ground is one method of reducing load transfer. As a result load transfer is reduced in both the longitudinal and lateral directions. Another method of reducing load transfer is by increasing the wheel spacings. Increasing the vehicle's wheelbase (length) reduces longitudinal load transfer while increasing the vehicle's track (width) reduces lateral load transfer. Most high performance automobiles are designed to sit as low as possible and usually have an extended wheelbase and track.
tropic gale
erm yeah I'm not too polished in this but this paragraph in the wiki might help ...
Yeah unfortunately not. Lowering is always a good idea yeah but for the spreading on the other axis it seems a lot more complticated
Like, if we did it in 2D
still surge
I have two identical cars. In each car, 4 weights of 50kg each are placed. One t...
i would say that spreading the masses would be the best
this keeps the wheels from leaving the ground
tropic gale
erm yeah I'm not too polished in this but this paragraph in the wiki might help ...
It's an extreme example, but there is going to be a definite difference between red and green
tropic gale
i would say that spreading the masses would be the best
Would it not be reverse? Here both reds are behind the front wheel
hardy lion
to be fair, the reds are also lower
still surge
in a four wheel example
tropic gale
reds?
The red weights. Made one image now instead of two. red weights is the scenario in which the weights are in the center and green weights are the ones where they are on the edges
still surge
if the masses are closer to each wheel, it allows the wheels to gain more traction
this makes a better turn
the effect of the centrifugal force is also less as the right fuselage mass balances the roll
since more torque and all
craggy sedge
i see, a fellow daoist
tropic gale
i see, a fellow daoist
Yep
craggy sedge
greetings
tropic gale
hello!
Well, I thank everyone for their help, but I guess it is a bit too detailed for a math server. Maybe physics too tbh, I will probably have to ask a car enthusiast community
.close
tulip yoke
is this correct
safe radishBOT
is this correct
hardy lion
indeed
craggy sedge
yes
tulip yoke
thanks
tulip yoke
is this correct
can this be solved using a different method? @hardy lion @craggy sedge
hardy lion
no
craggy sedge
kinda but not really a different method
tulip yoke
kinda but not really a different method
can u show? pls
craggy sedge
can u show? pls
$2^{y+3x}=e^{\ln 2^{y+3x}}$ and do it for 2^-3/2 but this is just the same thing in the end
flat frigateBOT
convergence
$2^{y+3x}=e^{\ln 2^{y+3x}}$ and do it for 2^-3/2 but this is just the same thing in the end
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.50 $2^{y+3x}=e^{\ln 2^{y+3x}}$ and do it for 2^
-3/2 but this is just the s...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```
river oriole
i mean, surely you just say the two powers have to be equal
so you just do power 1 = power 2
hoary anchor
Thats what i was thinking
craggy sedge
yeah it works for the reals ,but not exactly for all cases
river oriole
fair
vagrant yew
What is the answer for 2 a
safe radishBOT
What is the answer for 2 a
solar hazel
,rotate
flat frigateBOT
vagrant yew
I mean how to do 2b
honest perch
do you know what's 2^5
vagrant yew
32
vagrant yew
They r asking in full form not evaluating
honest perch
what's the answer to 2a?
solar hazel
mayervietorus
honest perch
:bending_skull:
vagrant yew
6x6x6x6x6
solar hazel
hello
honest perch
hey
vagrant yew
That is answer
honest perch
i'm guessing 2b is 1/(2x2x2x2x2) then? idk
edgy vine
hey can i get help with this
safe radishBOT
hey can i get help with this
edgy vine
the sketching part
i know it will get very small and go towards 0 when k gets very large
and its even
safe radishBOT
@edgy vine Has your question been resolved?
gusty inlet
you can calculate the derivative of the function and calculate its critical points, this should make graphing it easier
edgy vine
ok
what about roots
@gusty inlet
this doesnt really make sense to me
<@&286206848099549185>
gusty inlet
this doesnt really make sense to me
wait, did you calculate the derivative
gusty inlet
no that's quite fine as well !
gusty inlet
I also think you can maybe combine both the sine and cosine in one cosine
edgy vine
ah
edgy vine
ahh i see
ill do that
im not sure if i am doing this corretly
not done this in a while can u help
@gusty inlet
crystal ridge
only the graphing part right?
gusty inlet
<@713741736261713950>
you seem to be on the right track
edgy vine
ok cool
gusty inlet
sqrt(a^2 + b^2) is amplitude
edgy vine
yup
gusty inlet
and the arctan(b/a) is phase
edgy vine
cool
gusty inlet
then you can easily graph since you know the period and can calculate the first point at zero
edgy vine
wait , how do i know the period
gusty inlet
the part apart from the constant phase tells you that, it's kb and k is the variable
edgy vine
so my theta=arctan(-1/kb) is my phase shift
gusty inlet
which means the period is just $\frac{2\pi}{b}$
flat frigateBOT
Mohamed Mohsen
crystal ridge
yea exactly
This comes from the fact that if period of a function $y = f(x)$ is P then period of $y = f(ax+b)$ is $\frac{P}{|a|}$
flat frigateBOT
Bagchi234
edgy vine
wait is kb not its frequency
gusty inlet
b takes the role of angular frequency in analogy to time, kb acts as $\omega t$, and generally the period is $T=\frac{2\pi}{\omega}$
flat frigateBOT
Mohamed Mohsen
edgy vine
the part apart from the constant phase tells you that, it's kb and k is the vari...
did we get this from what we were already given? or from combinig the cos and sine
gusty inlet
did we get this from what we were already given? or from combinig the cos and si...
after you combine them you get this form $f(k) = A\cos{(kb+\phi)}$ which is similar to $f(t) = A\cos{(\omega t+\phi)}$ and from here you can get all the info you want, like $f(0) = A\cos{(\phi)}$ and $T = \frac{2\pi}{\omega}$
flat frigateBOT
Mohamed Mohsen
gusty inlet
actually i just noticed something extra, the amplitude changes with k too so you will have to graph the cosine shaped by that amplitude
but yes everything else checks out
edgy vine
cool
gusty inlet
also I wanna tell you something important
if you wanted to check the nuances of maybe a sign that could be wrong or anything you missed
you can graph the function before and after you changed it by desmos
edgy vine
ok cool thanks
gusty inlet
ok cool thanks
you are welcome ^ ^
edgy vine
wait, for this, as k gets bigger, this will mean that y gets bigger
cos the amplitude will get larger
but in the original equation wont
gusty inlet
how do you know it won't
there is a k multiplied by the cosine
it will shoot to the skies for high k
gusty inlet
the whole thing being zero implies either the amplitude is zero or the cosine is zero
you did the amp
what about the cosine ?
this will have infinite roots by the way
safe radishBOT
@edgy vine Has your question been resolved?
modern bloom
Given the following three-variable equation:
$\mathrm{xarctanx}+\frac{\mathrm{lnx}}{\mathrm{lny}}+z\mathrm{e}^{\mathrm{sinz}}=\frac\pi4$
According to the implicit function theorem, in a neighborhood of the point (l, e, 0), the equation can determine:
(A) Only one implicit function $z=z(x,y)$ with continuous partial derivatives.
(B) Two implicit functions$y=y\begin{pmatrix}x,z\end{pmatrix}$ and $z=z\begin{pmatrix}x,y\end{pmatrix}$ with continuous partial derivatives.
(C) Two implicit functions$x=x\begin{pmatrix}y,z\end{pmatrix}$ and $y=y\begin{pmatrix}x,z\end{pmatrix}$ with continuous partial derivatives.
safe radishBOT
Given the following three-variable equation:
$\mathrm{xarctanx}+\frac{\mathrm{ln...
flat frigateBOT
riyobi
safe radishBOT
@modern bloom Has your question been resolved?
safe radishBOT
@modern bloom Has your question been resolved?
safe radishBOT
@modern bloom Has your question been resolved?
fervent parcel
Hii, can someone help me graph this?
safe radishBOT
Hii, can someone help me graph this?
thin bridge
what have you tried
fervent parcel
I tried the first one already but i got a little bit confused with the 2nd one and 3rd
thin bridge
which part of those is confusing you
fervent parcel
Oh wait for the starting point for the 2nd, do you have to substitute 0 and also substitute 3/2 for the end point? Just only those 2 or i substitute other values between 0 and 3/2?
thin bridge
its linear so you only need two points to be able to draw it
(but remember to put a hole at x=3/2 for that part)
fervent parcel
Ahhh i thought of it in a complicated way 😅
For the starting point for the 3rd equation, is it (3/2, 17/4) ?
fervent parcel
I tried to substitute 3/2 to x at -x^2+2
thin bridge
can you show your steps
fervent parcel
My paper is a bit messy so ill just type how ill solve it.
=(-3/2)^2+2
=9/4+8/4 = 17/4
thin bridge
you're messing up the order of operations
fervent parcel
Oh is it the sign
thin bridge
the ^2 applies to the x only, in this case (3/2)
the - is applied after
fervent parcel
Ahhh so - 1/4
thin bridge
yes
fervent parcel
Though im not sure why when it comes to graphing, my teacher said that it supposed to be a curved slope
Nvm i think i get it
Thank you
.close
safe radishBOT
safe radishBOT
Available help channel!
edgy vine
safe radishBOT
edgy vine
can someone help with the very first bit of this
i hate trying to solve it when there is multiple moduli
soft matrix
i hate trying to solve it when there is multiple moduli
The graph or sol of the eq
edgy vine
both
eternal carbon
can someone help with the very first bit of this
solve the question in intervals where the moduli split, so consider x < -1, -1 < x < 2, and 2 < x
edgy vine
ive completly blanked on how to solve this
ok
what do u mean by solve it
ok what should i do after this
llike slowly step by step aha sorry
junior smelt
solve the question in intervals where the moduli split, so consider x < -1, -1 <...
In each of the listed cases here, you can write what the absolute values become
edgy vine
ok ok
solar hazel
In each of the listed cases here, you can write what the absolute values become
:waves:
edgy vine
but, like they will all be positive coz they are absolute values
im a little confused on what u mean
junior smelt
but, like they will all be positive coz they are absolute values
Have you seen the piecewise definition of the absolute value?
junior smelt
Do you mind stating/showing it, please? 
edgy vine
well i would recognise it if it was given in a question but ive only seen it for specific lik questions
maybe i need reminded
but its like y= something different for different values of x
or if x>something and x<something
but maybe i need reminded aha
is it when the function is given in those big brackets
cos thats what im thinking of
junior smelt
well, you have most of it, it’s
[
\abs{x} =
\begin{cases}
x & x \geq 0 \
-x & x < 0
\end{cases}
]
flat frigateBOT
@junior smelt
junior smelt
Typing on mobile
edgy vine
aha ok
junior smelt
Anyways yea basically you split it up depending on whether your insides are positive or negative
junior smelt
Basically for each of the cases, you use the definition on both of the absolute values
Consider whether the insides of the absolute values are positive or negative
junior smelt
Yep that’s the idea 
hexed igloo
junior smelt
Well, not quite yet
junior smelt
You need to put what those absolute values work out to be back into the equation, then solve what you end up finding
edgy vine
You need to put what those absolute values work out to be back into the equation...
wait so, i sub in like (x+1) then -(x+1) , but then what do i do with the opther absolut value
this is where i get confused
junior smelt
wait so, i sub in like (x+1) then -(x+1) , but then what do i do with the opther...
You work out the signs of it too, so like when x < -1, you have that |x + 1| = -(x + 1), but if x < -1, then x is also less than 2, soooooo...?
junior smelt
Yep, that's why there are the three cases, between -1 and 2, and greater than 2
junior smelt
Oh yea, was gonna mention as well, but one thing also is sanity checking the answers you find
You'll get linear equations when you break the absolute values up, but the solutions you find from those may not match up with the "actual" solutions, for example...
edgy vine
yeah of x>2
junior smelt
x=5/3
...you can end up finding values that aren't in the range under consideration, i.e. this one should be > 2 but it isn't 
junior smelt
Yep, discard it, it's not an actual solution to the equation 
edgy vine
nice
junior smelt
Then of course the other two cases you find, those are just left to do
flat frigateBOT
edgy vine
nice
edgy vine
why is there not an x=1
junior smelt
Hmmmm 
edgy vine
odd, did i do something wrong
junior smelt
Hmmm, nooo it seems to work actually 
Damn it, I just wanted to be lazy damn you Wolfie 
edgy vine
ahahah fr
junior smelt
I know right 
fathom jewel
\fbox{\parbox{\textwidth}{
\textbf{Problem.} \
\text{Find the limit or prove it doesn't exist of}
$$\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{x^8+y^{14}}$$
}}
$\$
\textbf{Solution.} $\$
\text{After some various tries, I came to the conclusion that it's probably} $L = 0.$ $\$
\text{The limit is weird so I will try to make use of the Epsilon-Delta limit definition}
\text{for the first time.} $\$
\text{Let} $\varepsilon > 0$ \text{and let} $\delta = \sqrt[10]{\varepsilon}.$ $\$
\text{Suppose } $0 < \sqrt{x^2+y^2} < \delta. $
\begin{align*}
\abs{ \frac{x^5y^5}{x^8+y^{14}} - 0} &= \frac{\abs{x^5y^5}}{x^8+y^{14}} \
&\leq \frac{x^4\abs{xy^5}}{x^8} \
&=\frac{\abs{xy^5}}{x^4} \
&\leq \frac{\abs{x}\abs{y^5}}{\abs{x}} \
&=\abs{y^5} \
&=\sqrt{y^{10}} \
&= \left ( \sqrt{y^2} \right )^{10} \
&\leq \left ( \sqrt{x^2+y^2} \right )^{10} \
&< \delta^{10} \
&= \left ( \sqrt[10]{\varepsilon} \right )^{10} \
&= \varepsilon \
\end{align*}
flat frigateBOT
𝔸dωn𝓲²s
eternal carbon
have you tried a path that makes the terms in the denominator the same power?
fathom jewel
yea
eternal carbon
what was the path?
vast obsidian
\fbox{\parbox{\textwidth}{
\textbf{Problem.} \\
\text{Find the limit or prove i...
At one point you are replacing x^4 by |x|. If we have |x| < 1 then then you make the demon bigger
fathom jewel
At one point you are replacing x^4 by |x|. If we have |x| < 1 then then you make...
\begin{align*}
\frac{\abs{xy^5}}{x^4} &= \frac{\sqrt{x^2} \cdot \left ( \sqrt{y^2} \right )^5}{\left ( \sqrt{x^2} \right )^4} \
&= \frac{\left ( \sqrt{y^2} \right )^5}{\left ( \sqrt{x^2} \right )^3} \
&\leq \frac{\left ( \sqrt{x^2+y^2} \right )^5}{\left ( \sqrt{x^2+y^2} \right )^3} \
&= \left ( \sqrt{x^2+y^2} \right )^2 \
&< \delta^2 \
&= \left ( \sqrt{\varepsilon} \right )^2 \
&= \varepsilon^2
\end{align*}
flat frigateBOT
𝔸dωn𝓲²s
vast obsidian
- to 3. Why does this need to be true?
fathom jewel
Ich habs gerade realisiert
fathom jewel
**𝔸dωn𝓲²s**
ja verkackt haste nh
vast obsidian
Ich habs gerade realisiert
Was passiert für y = x^(3/5)? Sollte dann nicht gegen 0 gehen. Schaue aber gerade Olympia, d.h bin nicht im Fokus
vast obsidian
ja verkackt haste nh
?
eternal carbon
what was the path?
.
fathom jewel
i have no idea i need to critically think now for a moment
vast obsidian
.
Sorry, I didn't understand immediately what you meant. Didn't mean to suggest again what was basically already said
fathom jewel
Sorry, I didn't understand immediately what you meant. Didn't mean to suggest ag...
can I go from x^8 + y^14 to x^2+y^2
but then i can not reduce at all
vast obsidian
can I go from x^8 + y^14 to x^2+y^2
No, they proposed x^(8/14) = y as a path
Both paths. Mine and his are working as a counter example
fathom jewel
and woher kommt das her
bruder ich hab so viel ausprobiert und jetzt gibt es die eine lösung
vast obsidian
and woher kommt das her
Rundprobieren würde ich sagen
Es war nicht einfach zu zeigen und deswegen ist die Aussage wahrscheinlich falsch
eternal carbon
Sorry, I didn't understand immediately what you meant. Didn't mean to suggest ag...
oh i c
fathom jewel
Kann man trotzdem mit epsilon delta es beweisen?
eternal carbon
all good
vast obsidian
Ich kann dir sagen wie ich auf meinem Weg gekommen bin: Ich wollte, dass y so wählen, dass du sowohl im Nenner als auch im Zähler x^8 hast
fathom jewel
ich habe versucht diese höhenlinie zu bestimmen
ich wusste es war iwas mit wurzel aber kamm nicht drauf
vast obsidian
Kann man trotzdem mit epsilon delta es beweisen?
Es ist ja nicht stetig
fathom jewel
ja stimm
diese definition is also dann geeignet wenn man weiß dass der grenzwert existiert
vast obsidian
Ja
fathom jewel
aber nicht als mittel um zu zeigen dass er nicht existiert, oder?
vast obsidian
Du müsstest halt zeigen, dass für jedes r in R je ein epsilon existiert, dass du kein Delta wählen kannst, dass diese Fehlerbedingung gilt
fathom jewel
alles klar
vast obsidian
Das ist schwer
vast obsidian
Deswegen einfach zwei Nullfolgen finden mit verschiedenen Grenzwert
vast obsidian
Eine Wahrheit die man nicht leugnen will
?
fathom jewel
Sag mal
???????????
reicht doch
danke für deine hilfe
bist krasser als du denkst
vast obsidian
Bitte
fathom jewel
einfach während der olympia schaut
probier mal x^8/14
😂
ich krieg nen schlaganfall
vast obsidian
Nah. Warum machst du die Aufgabe eigentlich?
fathom jewel
Nah. Warum machst du die Aufgabe eigentlich?
Der Grund:
#1272257901670563900 message
vast obsidian
Ah, ja. Habe mich schon gewundert. Sieht stark nach Ana 2 aus
fathom jewel
Ich hatte wie gesagt rumprobiert für eine Stunde das wieder analysiert recherchiert und dachte nach 3, 4 Wegen, das könnte gegen 0 gehen
Ja mein Ana2 ist einfach lächerlich
vom Niveau
Noch nie sowas mit Paths gemacht
vast obsidian
Ist? Dachte du bist im 4?
fathom jewel
Ich bin im 5. jetzt
trotzdem hole ich privat Sachen nach die ich nicht ander uni hatte
mein ana professor für ana 1 und 2 der hatte kein skript
vast obsidian
Och, ne
fathom jewel
wir haben so aufschriebe aus seinem kopf gemacht
obwohl der eig chillig war und gut erklären konnte realisiere jedes semester wie scheiße das war
vast obsidian
Sowas hasse ich. Ich schreibe nicht mit während der VL. Dann checke ich nichts
safe radishBOT
safe radishBOT
Available help channel!
normal harness
claim
normal harness
hi, is this the right answer
tardy mango
tardy mango
hi, is this the right answer
Yeah
You can check stuff like this yourself with wolfram alpha
normal harness
You can check stuff like this yourself with wolfram alpha
I dont think I have leanrt that yet
and thank you so much
also dumb question as its my last attempt, is this acceptable form?
tardy mango
I dont think I have leanrt that yet
Wolfram alpha is a bot lol
normal harness
I know it is but I wanna double check lol
tardy mango
Wolfram alpha is a bot lol
normal harness
Wolfram alpha is a bot lol
I thought its another theorem lmao
tardy mango
also dumb question as its my last attempt, is this acceptable form?
Yeah
tardy mango
I thought its another theorem lmao
I have to admit you make me chuckle a bit
tardy mango
I have to admit you make me chuckle a bit
First time I’ve heard that
normal harness
now no more attempt the question is locked hahaha
normal harness
First time I’ve heard that
its just end of semester paranoid
is it going to be my next chapter? ykyk
.closed
.solved
safe radishBOT
tardy mango
its just end of semester paranoid
^_^
safe radishBOT
Available help channel!
wide prairie
How do i simplify stuff like this with so many subtraction and addition signs?
I'm a newbie please teach me 
safe radishBOT
@wide prairie Has your question been resolved?
lean otter
safe radishBOT
lean otter
<@&286206848099549185>
gusty inlet
<@&286206848099549185>
you gotta ellaborate the question further for someone to be able to help
lean otter
you gotta ellaborate the question further for someone to be able to help
Find Pn,m
It is a function
In n and m
Can u pls help?
lean otter
you gotta ellaborate the question further for someone to be able to help
.
Can anyone pls help
@hel
<@&286206848099549185>
terse lichen
,rccw
flat frigateBOT
terse lichen
,rccw
flat frigateBOT
lean otter
lean otter
,rccw
.
Pls help
<@&286206848099549185>
<@&286206848099549185>
Yes
U need to find pn,m where pn,m is function of n amd m
Find it in terms of n and m
lean otter
Find solution of this function equation
It is the case of ballot problem
n and m are variable
They are independent
yes
Hwo to solve it
?
safe radishBOT
@lean otter Has your question been resolved?
edgy vine
safe radishBOT
edgy vine
hi can someone help me with the last true or false question here
ive forgotten how to do these
burnt notch
Hint: use implicit differentiation
edgy vine
ok
should i get -3x^2/(2y-6y^2)
im a bit rusty with this as these are quite old past papers im on aha
burnt notch
Hint: use implicit differentiation
Because you don't have y = something(x)
burnt notch
should i get -3x^2/(2y-6y^2)
Yep, that's correct
edgy vine
but there was an x, it goes to a one when differentiated?
burnt notch
Yes exactly
I suggest you to use y', instead of dy/dx, so that you can write shorter
burnt notch
You forgot to multiply by 2, when differentiating -2y³
burnt notch
But the rest is all right
safe radishBOT
Available help channel!
lusty parcel
lone hornet
Can you find the area of the triangle that is the base of the prism?
(assuming you need help with part a)
lusty parcel
i need help for both 😭🙏
lone hornet
if you need help with b you can just find h(t) or you can use dV/dt and dh/dV
lusty parcel
dont u need dh/dt too
lone hornet
you do need it in the sense that it's the answer for part b
nocturne cloak
safe radishBOT
safe radishBOT
@nocturne cloak Has your question been resolved?
desert swallow
x^2 = 29+20cos110degrees?
Yeah that works
nocturne cloak
Yeah that works
Ohh
Did i do anything wrong? All the answer choices is like 5.7 5.2 6.7 so i might be right or wrong
safe radishBOT
safe radishBOT
Available help channel!
golden loom
safe radishBOT
golden loom
is the answer 2log(2b)
magic junco
incorrect
golden loom
then how to do
magic junco
show me your work
magic junco
I can help you check where you have gone wrong
golden loom
quick crater
i dont think you can cross them?
golden loom
why
quick crater
alr, could you tell me how you got this
quick crater
that only works for $log(a)-log(b)$
flat frigateBOT
Skill_Issue
golden loom
so cuz of the -2 i cant?
quick crater
no its cause its log/log
golden loom
then how will i solve whatever inside that
how about log(a^2 - b^2) - 2log(2b)
then its not log/log
quick crater
$$\frac{log(a^2)}{log(b^2)}=log(a^2-b^2)$$, if you try to subtract 2log(2b) which is equal to log(4b^2) from it youll get
$$log(\frac{a^2-b^2}{4b^2})$$
flat frigateBOT
Skill_Issue
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
quick crater
yeah youll get something completely diffrent
also im pretty sure first step into second step is wrong lol
$log(a)\cdot log(b)\neq log(a+b)$, atleast not for all values a and n
flat frigateBOT
Skill_Issue
golden loom
but this is what i read
quick crater
?
cedar void
but this is what i read
it is only for $\log_z{a}+\log_z{b}=\log_z{ab}$
golden loom
whats wrong in step 1 to 2
flat frigateBOT
77²
golden loom
yeh
magic junco
Log isn’t necessary at all 😂
golden loom
the question is from logarithms
magic junco
Did you teacher demand you to apply log to this question?
golden loom
im learning on my own
golden loom
well i need to study log
is my solution all nuts or i can fix 1 or 2 steps and get the answer
cedar void
is my solution all nuts or i can fix 1 or 2 steps and get the answer
right hand side of first step is wrong
ig you should re-do it once
cedar void
can you tell me how can i write a^x in log
if $a^{x}=b{\implies}x=\log_a{b}$
flat frigateBOT
77²
cedar void
where is the base in xlog a
see in general you take the base of logarithm to be e
so log(a) doesn't become 1
cedar void
so if i have to change a^x in log it will just be log a^x
yeah
golden loom
wont the value change
cedar void
wont the value change
of what?
golden loom
a^x and log a^x are different right
cedar void
if you do it on both sides of any equation then no
golden loom
different values
golden loom
if you do it on both sides of any equation then no
ok
golden loom
right hand side of first step is wrong
how
golden loom
ok