#help-26
226100 messages · Page 252 of 227
Do you know what separation of variables is in differential equations?
um I don't think so
Well basically just put all the r’s
On the left
And all the t’s on the right
Which is almost what you did
But..
That thing you called b which is dr/dt has a t!
yup
yes, that makes sense
Let’s go back to this
Cuz what you did became messy
From this equation try and separate r and t
Note: k and a can be wherever they don’t matter
Not quite
oh yeah xD
What’s the integral of x/(1-x^2)
-ln(x^2-a^2)/2 +C
-ln(a^2-x^2) +C
Yea you don’t need to divide by 2 cuz you already have 2r
And the denominator is a2-x2 not the other way around
please can you explain this?
Here you should divide by 2 cuz I said integral of x/… not 2x/…
Well can you tell me how you got -ln(x^2-1)/2
oh yeah but we I have x^2-a^2 rather than a2-x2
let u = a2-x2
for some u sub
we get dx = -1/2x du
good
so we have -1/2 \int 1/u du
ok
now before you finish
do the same steps for the integral we care about
integral of 2r/(a^2-r^2)
exactly
yet u = a^2 - x^2
yup
so we have -ln(a^2 - x^2)
+c
ah yes
so there you go
so kt = -ln(a^2 -r^2) + C
the radius
yup
indeed
perfect
so for these type of problems we should seperate variables, then integrate, then use conditions?
exactly
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Who knows how to tackle this -1 into a logarithm?
is that a "minus 1" ?
Yes
you don't need to. just add 1 to both sides
and what equation are you working with and what variable are you solving for?
use $\log_2(2) = 1$
riemann
What?
,w log base 2 of 2
what are you confused by?
yes
#help-4 message
it's also much more standard in the math resources as well.
than $^g \log(a)$
riemann
I’m sorry I’ve just been taught differently hence the confusion
I just want to know how to get the -1 into the logarithmic
I have no formula’s
do you understand this now?
Yes
put it in here
I come to ^2log(x^3) -1
But -1 needs to be manipulated into the logarithm, that’s my question
riemann
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^2log(2) = 4 not 1?
62 students are picking two activities to do over the weekend.9 picked painting and sport.
11 did not pick painting or sport.
Twice as many students picked sport than painting as one of their activities.Find the amount that picked sport and not painting.
Its 1
How is 2^2=1 and not 4?
Let log (base 2) 2 = k
Then 2^k=2^1
so k=1
So log (base 2)2=1
I can only assume that ^g log(a) isn’t the same as the above mentioned
It’s also looking like Chinese to me
Perhaps I should go sleep 💤
Is the answer 31
do you have any calculators? this is a basic property of logarithms
like i said just get used to the notation that a majority of people who use math use. sorry if you're in the minority, but it's not that complicated to remember
@sturdy vigil Has your question been resolved?
yeah but idk how
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Could i have help
S(y2-y1)dx =576 then solve for c.
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for p-series, why arent they convergent at p = 1
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Hi I need help solving the scale factor
When I solve it the way my teacher showed, I get an answer of 2
but when I do it in reverse, I get 0.5 which is what the textbook says
AB and DE are corresponding, to get from m(AB) to m(DE) what do you have to scale AB by?
2
You got it correct
1/2 is the correct answer
so then why does the textbook say 0.5
DEF is the image of ABC
Because the textbook is correct?
The answer is 1/2
2/4 = 0.5
AB * k = DE
8k = 4
k = 4/8 or 0.5
so smaller triangle to larger triangle? basically instead of 8/4 I do 4/8
the way i solve it is:
n = AB/DE = number/number = number
but instead of AB/DE
do DE/AB?
nvm ur way actually makes a lot more sense
thanks man
much appreciated
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wait a minute @junior widget
can u explain how my teacher got this then
scale factor of 2
i just did it ur way and its also 0.5
is her method even corrrect
@junior widget
Your teacher doesn't understand why they are using ABC and DEF
DEF is the image of ABC
At least, thats how the alphabet works
Either way, your answer and your teachers are both Correct in a way
thats helpful but also not helpful cause idk wth shes gonna quiz me on when the time comes, gotta ask her tmrw
thanks for the help man
much appreciated
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use the sine rule.
im grade 10 bro idek what that is
nvm
a you know trig?
just started it today
oh crap
you could google the cosine rule
it will help you solve it in a second lol
we're prob gonna learn it cause my teacher made a bunch of sub folders
im guessing its the same with the quadratic functions cause our teacher made us go through hell before showing it to us
factoring is not too bad
factoring is easy
getting the roots without the quadratic equation in a trinomial is hard
especially if ur me
use a computer lol
i used to use some wolfram alpha widget
it's much easier if you the computer and get the answer then factor it to get the answer
i think ur teacher got the y wrong
cause of the scale factor?
Why would it be wrong?
the scale factor is questionable but everything else seems right
shes subbing it in from the other shape
It's similar shapes, so $\frac{12}{6} = \frac{y}{4.5}$
dldh06
So y = 9
i tried out the cosine rule and it give me 4 sqrt 7
oh yea similiar shape
cant wait till i learn the cosine rule
:D
anyways thanks for all the help
u guys have a great night
😘
cya
FYI, your channel so you have to close it
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Can someone help me with this proof
@gleaming pasture Has your question been resolved?
what have u tried
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anyone expert in algorithmic graph theory
please give a shout. need help
@void edge is it a very long/complicated question?
You might wanna ask in #numerical-analysis
I’ve seen some questions that are very long
It would be easier to find someone who’s willing to read beforehand
Cause most people see big paragraph and nope out
but i just needed an excuse to post that sticker
lol
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Hey, what is the correct answer to the integral of 1/(3x) dx?
I got 1/3*ln|3x| when I used the formula 1/(ax+b) dx = 1/3*ln|ax+b| and my teacher got 1/3*ln|x| when he separated 1/3 from the original integral and integrated 1/x alone since constants survive.
Bottom line my teacher thinks we’re both right but he wasnt able to prove that the results were the same by trying to get to 1/3*ln|x| from 1/3*ln|3x|.
If you derivate both integrals you do get back to 1/(3x) 
derivate both integrals
differentiate, not derivative
wasn't able to prove
But you differentiated both results and showed that they're the same?
The key is to realise that antiderivatives differ by a constant. 1/3 ln|x| - 1/3 ln|3x| = 1/3 ln|x/3x| = 1/3 ln(1/3) which is a constant
Ah we use derivate also as a verb in Norwegian, we “derivate” to find the derivative
Sure, in Norwegian, but not in English
I didnt learn math in English 
Yes, well, I differentiated both integrals in my head after class but he was trying to get to my result using his result and some logarithmic formulas… this was actually part of an answer to a longer integral, but this was the only difference in our answers
I’ll share both our answers when I get home
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@sterile ridge Has your question been resolved?
Hey, so I'm back home, and the original math problem was; ∫(x² + 2x + 1/(3x)) dx where I got 1/3⋅x³ + x² + 1/3⋅ln|3x| + C and my teacher got 1/3⋅x³ + x² + 1/3⋅ln|x| + C
My teacher thinks we're both right, but struggled proving it. While we can see that these are two different solutions, how can these be the same? And if they're not the same, which is correct?
So with this
The key is to realise that antiderivatives differ by a constant. 1/3 ln|x| - 1/3 ln|3x| = 1/3 ln|x/3x| = 1/3 ln(1/3) which is a constant
1/3ln(1/3) is part of the constant C?
Thing is though, if you inserted a value for x, wouldn't these two functions produce two different results?
So because of this, these can't both be correct solutions..?
No.
For example, integrate 1/(3x) dx from 1 to 3.
Use 1/3 ln|x| as antiderivative to get 1/3 ln(3) - 1/3 ln 1 = 1/3 ln(3).
Use 1/3 ln|3x| as antiderivative to get 1/3 ln 9 - 1/3 ln 3 =1/3 ln(3).
Any constants cancel when doing the integration
So, since C is an unknown constant, we can't for sure know what f(x) will be if we integrate f'(x), thus f(x) will differ by C which includes 1/3ln|3x|, making both integrations correct?
Is that what you mean?
$\int \frac{1}{3x}{dx}= \frac{1}{3}\int \frac{1}{x}{dx}$
Umbraleviathan
But then I'm getting (1/3)ln|x|
Same as my teacher
It can't be ln|3x| because by chain rule, if you derive ln|3x| you get (1/x)
However, if you use the formula ∫1/(ax+b)dx = 1/a⋅ln|ax+b| + C, then you get my result
where a = 3 and b = 0?
Uh lemme think
No that's not right
Lemme think
Oh you know what
That only works when b ≠ 0
That's because if b = 0, the "a" would cancel out
Leaving you, by chain rule, with (1/x)
Chain rule
(1/3 ⋅ ln|3x|)' = (1/3 ⋅ ln|u|)' = 1/3 ⋅ 1/u ⋅ u' = 1/3 ⋅ 1/(3x) ⋅ 3 = 1/(3x)
No 🙃
If we differentiate both integrals, we get back to where we started
So both have to be correct, unless I'm missing something
Oh atcually
Hmm
Well m $\frac{1}{3}\ln{|3x|} = \frac{1}{3}ln{|x|} + \frac{1}{3}\ln{(3)}$
Umbraleviathan
Which would get the same result if you derived it
True, so, can one say that 1/3⋅ln3 is part of the unknown constant C?
You could, but it isn't the value of C
No, but it's part of it, right?
Mmhm
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The only way I think you would've gotten your answer is by u-subbing then
3x = u
This is how you get (1/3)ln|3x|
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Is there a way to find the range just from the equation? or do you have to draw the graph and points
the 3 in the equation represents the y value so I get the 3, but how do you know whether it's going to go to positive or negative infinity from just the equation?
There is a way to find the asymptote
Then you know it keeps going to to infinity
The asymptote is just 3 because the +3
How do you know it keeps going to infinity?
Because it's an exponent
All real values work
So you can just keep making x larger
,w graph 2^x + 3
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what's the latex of
$$f(x) = a\log_b(x-h)+k$$ do you want those arrows too?
ohNoiAmHere
No, thanks
Btw, the b of the logarithmic graph doesn't represent growth / decay like how the exponentials does right?
hmm, it's simply the base of the logarithm, something with a larger base will be smaller. its not really growth though
there are models with log though
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do you guys know if these two equations represent the same plane?
-x+3y-z=0
-5x+15y-5z=0
I guess it does since u can multiply the top equation with 5
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can someone help me please
First you'll want to find x in terms of θ
Then the calculus is the easy part. Differentiate in terms of t, keeping in mind that dθ/dt = π
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what is half of 10 powered to 500? And what number gets the same result if you multiply it by 0,2 or subtract 0,2 from it?
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hi
i'm curently tring to do the 1 (a) for this i need to modifiate the 3 expretion to get the 1 ou the 2
for the moment i try partial integration but i don't fin an end to the cycle, i have been thinking of using change of variable but can't find one usefull so if you have any lead please
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Also see the given figure of thos above question
I don't understand that whether they are talking about area of whole triangle or right angled triangle .
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help
im stuck
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i need help with a decent amount of questions if anyone can help pls do i have a big exam tmrw
its to late
do you know about vector addition?
yes
whats the angle between those two vectors?
90?
no need for that
what did you take as base?
not tan-1(theta) i mean tan-1(15/5)
You did a common mistake
nope
you've taken 5 as the base
so the angle you got is wrt east direction
got what you did wrong?
draw the theata; there are two angles here and you're getting confused b/w em,
you're getting confused b/w these two angles
perfect, so the angle is 71.6° right?
yes
now look at the options
perfect
did you know why you did that?
because in the options you're trying to find the angle wrt south which is y axis, whereas the angle you calculated is wrt x axis
yes
because you've taken 5 as the base and tan theata = perpendicular/base
perfect
so if you've calculated tan-1(5/15) you'll get the angle from y axis
@timber jungle got it?
i got the first part but i dont rlly get why i would do tan-1(5/15) can u show me a triangle with the xy plane pls
this may be a dumb question but how do i know which is the perpendicular and which is the base ,-,
I got the same question when I started trg.
1 min I'll explain
@timber jungle read it few times and try to visualise what I'm saying you'll get it ;)
the last lines are enough to get it
alr i get it i think
7?
fuck this isn't a right angle
ye was gonna ask that xD
assume the angle to be 90°
"the line opposite to theata is the perpendicular" you need some visualisation
check some yt videos it'll help
isnt 7 the line opp to the angle?
so 7 is the perpendicular
hence 9 is the base
so you got it right?
ye ye i get it
nice
look at the problem, now you'll be able to solve it
tan theata = (perpendicular/base)
np!
have my physics work pending
<@&286206848099549185>
plug in the polar to cartesian transformations to the polar equation
r^2 = x^2 + y^2 etc
i dont get it i tried multiplying the whole thing by r
first step is $r (1 + \cos(\theta)) = 2$ then use $x = r\cos(\theta)$ and/or $y=r\sin(\theta)$
riemann
you want to substitute all r and theta for x and y
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@sweet shard
Please don't occupy multiple help channels.
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Yeah I don't know exactly what it means
You don't know what the chain rule is?
I know what the chain rule is
-_-
A composite within a composite
derivOuter • derivInner • derivInnerInner • derivInnerInnerInner ...
So let's start with the sine
Take the derivative with respect to the outer
What would that look like
Yeah so that's a beginning
So now you gotta multiply that by the derivative of sin(6x^7)
Doing this step-by-step
$$\frac{d}{dx} f(g(h(x))) = f'(g(h(x))) \cdot \frac{d}{dx} g(h(x))$$
just as a note, you can think of g(h(x)) as a single function and use chain rule as if you had f(g(x)) then for the derivative of g(h(x)) you apply chain rule again
Doggo
why is that
I did a similar problem
i'm not trying to be difficult i'm just trying to figure it out so i can do it myself later
Yeah you can use J42's as an example
You gotta like
Well you have the first step done
I would think that would only apply if sin^2(x) = sin(sin(x))
brb gotta grab something from the oven
Umbraleviathan
First let u=6x^7 and find du/dx
It's easier to just use chain rule for this
Then let v=sin(u) and find dv/du
Finally differentiate your term with respect to v
Then multiply by du/dx and dv/du
Its the same thing but more working
Just watch black pen red pen on YouTube and you will be brilliant at differentiation it really helped me this year in maths
so uhh then you get
cos(6x^7) right?
or no
I forget how to take a derivative of sin with something other than just x in it
is that what you mean
oh wait am I supposed to use chain rule here
like
$\frac{d}{dx}(sin(6x^7)) = cos(6x^7)*(42x^6)$
Eyesonjune
Yeah
now what do I do
$\1+1
Eyesonjune
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I'm very stuck at b)
can you help me understand the hint? I'm not sure what it means by applying T n-1 times
for (b)
$T^{n-1}(v) =T(T(....(T(v))))$
riemann
for n = 3: $T^2(v) = T(Tv)$
riemann
If $v$ is an eigenvector, then $Tv = \lambda v$
riemann
Since $T$ is a linear operator, factor out lambda
riemann
I'm not sure how this proves linear independence, do I have to factor out lambda n times for every v_n?
do the hint first
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what does a discriminant show?
$b^2-4ac$
yomiko
it's one of the best way to determine the number of roots of a quadratic
does it only work on x^2 function?
yes
i mentionned "quadratic" so yes
what about if i want to know how many roots on other functions
like x^3 or x^4
^
A polynomial of degree n has at least n roots
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Hello I’m very stuck at this question I don’t get it
This is basically my work
But it still feels wrong 🫤
Not sure if I’m doing it correctly but I cant grasp what the videos are saying either
hello, the key thing to notice is that triangles BAC and DBC are similar (they share angle C, they both have a 90 degree angle, therefore all their angles are congruent)
Using ratios of corresponding sides of similar triangles you can see there is this relation:
4/x = x/3; solving for x yields x = 2√ 3
here are the corresponding sides of these similar triangles if that makes it easier to visualize
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Thank u so sm ❤️❤️❤️❤️❤️ this made it so much clearer
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I'm trying to get the primitive function of:
The answer is supposed to be
But I have no idea how that could be the answer
what I tried was just
primitive function is the antiderivative?
Yeah
ok
do you recall the rules for derivation when $f=e^{ux}$ ?
madlor
ok, whats $f'$?
madlor
that is correct
okay
now do you agree that the antiderivative would be the excact opposite?
I tried doing that but in reverse
yeah, what did that look like?
go for it 👍
yep I was just overthinking things
All I had to do was take 150/-0.3
ty for the help
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"A solution contains 1% salt. The solution is set outside and aftersome time, water loss has occured. The solution has lost x% of it's mass and the new salt content is 10%. What is the value of x?"
I setup two equations, $\frac{S}{m} = 0.01$ which describes its intitial state. After the loss, $\frac{S}{mx} = 0.1$ We can subsitute S from the first equation to the second one: $\frac{0.01m}{mx} = 0.1 \Rightarrow x =\frac{0.01}{0.1} = 0.1$
Duck Observer
The answer sas "x has to be greater than 80%"
Which means mine is wrong
Could anyone guide me?
losing x% mass means you multiply by $1 - x$ assuming $0 \leq x \leq 1$
Doggo
@neon iron
Wait wait
Does that mean that the final answer is umm
$x_f = 1-x = 1-0.1 = 0.9?$
Duck Observer
Ah
I'm not sure if that'd work but I doubt
your equation is wrong
I will try and solve it with (1-x)
Oh 😦
Solving it with (1-x) in the second equation gives: $\frac{0.01}{(1-x)} = 0.1 \Rightarrow x = \frac{0.1-0.01}{0.1} = 0.9$
Duck Observer
mhm seems right
Thank you, may the Duck Bless you, Doggo
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Hey, can someone help me find the derivative of equations of type x^2 + y^2 = 1?
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how does induction work with the special E that means the sum of all?
llike for example this one
when i get to the step of replacing N with K do I just leave the I's alone? or do othey get replaced with k's too?
I don't need to solve it fully I only need to get it to the form of what the left and right side would be at k+1
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Im a bit confused about the K+1 step as well Like I understand your just putting K+1 instead of K but theres another thing you have to do to it that I'm not seeing and I don't quite get what I'm missing about it.
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can someone explain how those two sets are equal
so C is the subsets of A
which is S with one more element
how do they have the same number of elements when C has one more element

C is not all the subsets of A
i meant B is
its the set of subsets that contain element d
my bad
it is the set of subsets of A
A without d
isnt it?
I mean it even says that
yeah its the set of subsets of A\ {d}
then how is | B | = | C |
for each set in B you can match a set in C, (one to one correspondance)
forall X in B it corresponds to XU{d} in C
and forall X in C, since d is in X
X = {d} U Y where Y is a subset of A\ {d}
so Y is in B
so there is indeed a one to one correspondance between elements of B and C
so the subsets of B are the subsets of C
but with d
hmm i think i got it, thanks
im gonna wait to see if anyone else has anything to add
if not ill close it
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for legendre's formula what if the number inside the brackets doesn't have a factorial?
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@fluid lark Has your question been resolved?
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You should show the formula and exactly what you mean
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Just doing some graphing functions revision, not sure how to go about rewriting this function\
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Rœmer
We can find $\frac{x^2 +x}{x-1} = \frac{x^2-1+x+1}{x-1} = \frac{x^2 -1}{x-1} + \frac{x+1}{x-1}$
and then rewriting should be fairly easy
Rœmer
I can see where you're going with it 
have you solved it ?
Can anyone help me with some homework problems
ask in one of the openhelp channels
nice !
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Thanks mate
np
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Umm I just asked a question but uhh I’m weak sry ;-;
This equals this ?? I’m not sure about it but got nothing in my mind I could do to solve it xd
,rccw
iCaird
,rccw
take a screenshot of it on the computer/phone
alright
and do they want every solution? just integer ones?
because with the way you've wrote it, finding all solutions is very very tricky and I don't think you'd be expected to find those
but if its meant to be what I wrote earlier than its a lot easier
so the problem just asks for roots in that interval?
Dunno how to say it in English xd
Like when the function come across x and y xd
Uhh
right, that wouldnt be touching
touching here would imply the curve is tangent to these axis
a better word would be "intercepting" or simply "crossing"
anyway, since you are restricted to an interval, it's easier to solve
what have you tried?
I tried doing this but not even sure if it works
Since they don’t have same angle
First 1 got cos x second got cos x^3
you work is wrong
cos(x) if not a factor of cos(x^3)
Wt u suggest xd
ok so we are finding the x-intercepts first, lets agree on this first ok?
I’ve been studying for programming exam for weeks so I’ve forgot a lot of math 😅
so we dont get confused and think about the y-intercept
anyway, notice that since cos(x)-cos(x^3)=0, that implied cos(x)=cos(x^3)
can you see how we can advance?
yes, but that's not what i meant
if cos(x)=cos(x^3), would it imply that x=x^3?
Not sure tbh xd
yes it would (kind of), because cosine here is a function, and a function (assuming that it is injective) can only be equal to itself if their input are equal to each other
or well another way of saying it is if f(x_1)=f(x_2) then x_1=x_2 considering that f(x) is injective
though, that's the easy part
the hard part here is cosine is not injective
What injective mean xd
one-to-one, one input going in and only one output going out
i dont want to go into this too much
But I understand this part
but basically just remember this, since cos(x) is an even function and periodic, hence if we have cos(x)=cos(y) then x=y+2kpi or x=-y+2kpi where k is an integer
im just saying those hoping that you would understand, what i want you to know is the message above
Isn’t it 2pi * k ?
(basically, i cant just throw you some random knowledge without explaining)
that....is the same thing
but anyway, now that you have this knowledge, could you solve cos(x)=cos(x^3)?
According to wt u write here xd
(note that since we are restricted to [-90,90], there is no need for the 2kpi)
the 2kpi here just represents the periodicity of the function, basically saying that there would be infinite solutions in general
Mhm
so anyway, it simplifies down to x=x^3 and x=-x^3
can you solve for x in these equations?
All I can think of is x = 0 xd or x = 1
Nope
ok, but one thing to notice here, x=1 is actually not in degrees
it is in radians
so you have to convert to degrees here for the answer to be correct
you are in radians, so you need to divide by radians, when you divide by radians you get the ratio between 1 and what you divided by, so you have to multiply by the same amount of what you divided by in degrees
hence 1*180/pi
Aha
Ok listen
This is the drawing
The 1 and -1 is the 180 in radians ?
Umm
Wait
Hold up xd
No Brian working atm
Those r the 1 we found
I feel like I lost my math knowledge…
Or our math teacher send question not normal
Cuz if I have to do all that in exam I’m finished /:
But how here they cross x= 0 too ?
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Hello I’ll translate xd
@jade cobalt Has your question been resolved?
Should I just close it ?
that's because there is a removable discontinuity there at x=0
and sorry, had to do something
the graph cross the x axis at (0,0) because the limit as x approaches to 0 of the function approaches the value 0
you probably dont need this knowledge now
and also yes, x=-1 is a root too, i forgot to mention that
x=x^3 gives 3 solutions
and what have you tried?
@jade cobalt Has your question been resolved?
I find out thx xd
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Hi, not sure how to find this
This is what I came up for the first part
Since now p_0 is constant and the derivative should be constant too
Does that mean I can just do it like this?
Seems like it’s not actually correct
I just made a simple linear demand curve and it does hold up, not sure if it works for a non-linear one, but this might be due to that assumption which I forgot the name of
Something like this but not exactly
Seems to hold up for non linear too
So turns out that is correct but would be really helpful if someone could guide me on why it’s like this?
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Find all functions that satisfy $$F(x) = \frac{x}{2} + xF(x+1).$$
ohNoiAmHere
i have no clue how to start this
i tried a substitution but got some ugly recurence
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@floral rampart Has your question been resolved?
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In line AB there are 4 segments ABCD A(1,2) D(4,8) AB=BC=CD find B and C by using midpoint formula
@long trellis Has your question been resolved?
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Midpoint formula
Just an average
Thing of it as an avaerage
Very simple formula
the midpoint formula isn't quite what you want to use here
But the questions asks you to use it
start by reading the question first @restive inlet
i am aware of what they wrote
sometimes questions are ill formed and/or it might be something they assumed they were supposed to use
This is what I mean sorry for my bad English
You don't have to spam sorry for my bad englihs all the time
Can u send the photo not sideways pleas
,rcw
Midpoint of ab?
