#help-42
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5x^2+5xn+12x/n+12n/n
and then identify the x terms, factor x out
k will be the coefficient
for what I mentioned earlier
and it should be easy to see that if non-zero integer |x| <= 12,
k would be relatively small
why 12 for n now
wasnt it 5?
because 0 < 12/x < 1 for values greater than 12
you'd also want to ensure there aren't any issues for lower integers
if there wasn't an restriction on values integers, there would be
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Can someone please help me format this LaTex doc? I need to get a picture to right of the text.
.close
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is this ap stat?
@lucid knot Has your question been resolved?
sorry for the late response, no it is not! this is algebra allegedly
its a statistics module in an algebra course to be specific @ionic moon
The mean is 8...Since the mean (7)
is 7 the mean or the median?
@lucid knot Has your question been resolved?
ahhh! good catch, the mean is 8 since the median is 7
the data is skewed lower, 1 is the lower extreme and 20 is the higher extreme but most of the data points sit between 5 and 11. The mean is 8, since it is right between 11 and 5. The mean is close to the median, so the data is somewhat symmetrical. i had sort of a brain moment, this should be better, does my reasoning stand?
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hello
ℝαμΩℕωⅤ
,, \int \frac{u}{(u^2 + 1)^2} du = \int \frac{2u}{2(u^2 + 1)^2} du = \frac{1}{2} \int \frac{1}{t^2} dt = \frac{1}{2} \left[\frac{-1}{t} \right] dt = \frac{1}{2} [-t]^{-1} dt = \frac{-1}{2} [u^2 + 1]^-1
oh, the last equality got cut off
higher!
wait so its the integral of this one ?
yeah
okay but why is it -1/2
see here
we multiplied the integrand by 1 = 2/2
that's all
we need the 2 in the numerator because dt = 2u
and then we pull out a 1/2?
yeah
but how exactly does that even happen tho cuz isnt 2/2 = 1
sorry
im slow today
LOL
no worries haha
you're right!
we can multiply by 2/2 because it's just equal to 1
but the 2 in the numerator combines with the u there to make dt
while the 2 in the denominator comes out as 1/2
nono
dt/du = 2u
oh
so dt = 2u du
that is so annoying
and (u^2 + 1)^2 is just t^2
because 2u du is dt, the 1/2 comes out, and (u^2 + 1)^2 = t^2
oh
you changed the question
like here wait
the antiderivative of 1/t^2 is -1/t
this follows from the reverse power rule
oh yeah
cauise
sorry i just get confused when i cannot do it step by step but i see it now
no worries
yeah
becomes the final answer
OKAYYYYY
okay now it is time to tackle the ohter Half
so i have this formula here
my handwriting is bad but
this formula somewhat applies but then again where did this come from
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help
how do i solve lg 2 using lg 5 = 0.699 and lg 6 = 0.778
lg 5 is lg(10/2)
Bro tried to confuse us by giving log 6
i have to use both ☹️☹️☹️☹️
its said in the question
should i do lg 2 = lg(5*2) = lg(10/5)?
my bad im new to lg so im sorry
The second expression is wrong.
2 is not equal to 5*2
but 10/5
awh man im stuck
in that case no
i gotta choose my fault
id choose lg 5 to find lg 2 right, but then what
frick maybe i should use lg 6
2 = 10/5
log 2 = log (10/5)
log 2 = log 10 - log 5
log 2 = 1 - log 5
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yeah
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Help
isnt it cosine?
draw out the vectors maybe
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it says this is wrong but my thought process was that all the tension forces are the same since they go the same amount from the origin with the same angle
so for the vertical components, T sin45 + T sin45 + T sin45 + T sin45 = 275*9.8
but idk what's wrong
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<@&286206848099549185>
@shell token Has your question been resolved?
so each rope has a component in the z axis, and a component in the xy plane. I'd start by finding the lengths of this triangle.
@shell token assuming you're still around XD
you can use 1/4 the mass of the block, because there's 4 ropes carrying equal tension
I'm saying you can model the each rope as two components, one pulling directly upwards, and the other pulling outwards from the center of the ceiling
so the tension in the vertical component is due to 1/4 of the weight. the actual tension in the rope is larger than that because it's not pulling directly up.
make sense?
We can calculate the tension in four hypothetical vertical ropes (it's just 1/4 the force requried to keep the mass still). And we can calculate a relation between the force in a vertical rope vs the force in the actual rope (using that cosine).
I've made a mistake there, it should be 5 / cos(t)
okay..
sry I was on call with my friend
he figured it out but in a completely diff way 😭
and im just more lost
but he did 0 = W - 4 (T1 * 5/sqrt(75)) and solved for T1 and it was surprisingly right
all I know is 5/sqrt75 is the unit vector or something
but I don't really get anything else
but sorry @rapid cliff didn't mean to ignore what you said, ill be glad to hear how you would do it now, im just trying to understand both ways
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what does it mean for a point to be on the circle
To verify the circle's equation
what does that mean
oh
i see
DF = PB sin B
DE = PC sin (A+B)
how do i show this is equal
Hum
I feel like its not asking u to prove this equality
But to prove the equivalent relation
but wont showing the above show that
Like assume equality is true, and show that implies the circle thing
k
Assume the circle thing, and show that implies the equality
k
but what result would lead to the point bieng on the apollonius cirlce
oh
done
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I thought it's called Euler's Notation for the last column (blue)?
And where is Partial Derivative Notation ∂f / ∂x
I only see these two in my Math (Calc 2) university class
I wish it was only LaGrange, but maybe it's not possible to teach all of Calc 2 using just this notation? Leibniz is mandatory?
i dont think its mandatory
@noble cedar Has your question been resolved?
also apparently arbogast notation is more historically accurate
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Fix bot sir
so as we know that cot theta = 1/tan theta
find tan (-0.003) first and then find it's reciprocal
as the theta is very small ,tan theta is approximately equal to theta which means
tan (-0.003) = -0.003
now for cot (-0.003) as I said we have to do 1/-0.003 after which we get approx -333.33
Im guessing itll be the same for
cot(3raiden/2)
find the tan of that then use its reciprocal
yes
tan(3radian/2) is undefined so will be the same for cot too
I hope your doubts are resolved
tan (3radian/2) is -1/0 and as the denominator is zero, it's undefined
it's complex, it could have been done more easily I think
for ex that cos 225 can be written as cos ( 180 + 45 ) and then we can solve it using cos ( a + b ) = cos(A).cos(B) - sin(A).sin(B)
If that graphical method is easier for you then you can go with it
I was taught this in my school
You in America?
I am from India
i have 40 mins ot turn iin these two quizez bro im cooked
I never thought someone would try to go for it graphically
so things are diff there as I see
undefined
So this is undefined too then. Cause cos is = to 0 and for sec it’s
Sec A = 1/cos
it's not bro .
sec ( 19π/2) can be written as sec ( 9π + π/2 ) = sec π/2 = undefined
yes
refer to this table
Okay bet
It’s hard using this compared to what I’ve learned in class but your way seems easier for sure
what grade are you in
coz that table is like the foundation of whole trignometry
the first thing to be taught and learnt
damn
American math is so stupid and all over the place
I understand the numbers so well. But the concept i do not understand .
well the way you are being taught even the simplest things in the most complex way
it explains
bro thats what im saying
my professor is doing the most dude. Its so annoying. Trig isnt suppose to be this difficult
you can remember that table, it will make things a lot easier
im just being told radians everynow and then and some numbers but not how everything comes together and when we use things
Bet thanks bro ima rmb the table. Is it okay if I add you?
sure np
@echo mirage Has your question been resolved?
what if csc is at 240?
cosec theta = 1/sin theta
so i make the 1=240?
sin (240) = sin ( 180 + 60 ) = sin 60
sin 60 = √3/2
so sin 240 is (- √3/2)
doing again that reciprocal will yeild
cosec 240 = -2/√3
ohh because it resets cuases its 2 rads in a unit circle
yes
okay bet that makes sense
it is an identity though
which is used majorly in these type of ques
based on the graphical method you use
on rationalisation we get -2√3/3
Flex man ngl
i think im cooked man
My brain is fried and all i have left is 4 questions
but i cant compute my brain anymore. my professor drop this quiz last minute it was 2 of them
well it is not smth you can just learn in few minutes and use it like pro
I will try
no -2√3 whole is divided by 3
bro I could be making errors
give me some time
not sure about the 8
8 is -√2
9 is 1/√3
Okay bet I got that at first
it is rationalised form of -2/√3
8 is -√2
9 is 1/√3
10 is 0
11 is -2√3/3
12 undefined
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A bit stuck
I tried to do byparts on the bottom left and then tried to continue it on the right page but doesn’t seem right
And a u sub wouldnt really help so im kinda confused
(The weird squiggly letter is supposed to be a “t”)
this is a first order linear IVP
if we just wanna know the "interval of validity" then theres a theorem out there that tells us what it is without finding an explicit solution
see theorem 1: http://tutorial-math.wip.lamar.edu/Classes/DE/IoV.aspx
In this section we will give an in depth look at intervals of validity as well as an answer to the existence and uniqueness question for first order differential equations.
is it (-inf,-2) (-2,2) (2,inf) ?
did you read the link? can you tell me how to apply it here?
yes, we have 5/(t^2-4) we look for something that makes the function = 0
so we have +2 and -2
and we're given y(-1) in the problem so we need an interval that contains -1, I think its -1<t<-2
the ln(t^2) will always be defined since squaring a negative will give a positive
ok thats one function, whats the other one to consider?
you should check this
theres another discontinuity you forgot
ah its also undefined at 0/any t
now whats the "t0" in this problem?
and the given initial value is y(-1) so we have to include -1 in the interval
so -2<t<0 ?
ok yeah t0=-1 and the interval containing t0 is (-2,0)
thats all! no need for explicit solution, isnt it neat?
yep thanks
was trying to integrate a monstrosity that couldnt be integrated for the past hour
at some point I gave up and knew it was a trick question lol
yeah I forgot to divide the ln(t^2)
not sure if itll get easier once you correct it
I think it would've still been pretty bad but not too sure
,w (t^2-4)y'+5y=ln(t^2)
for sure you need to use the theorem instead of trying an explicit sol 🙂
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no prob!
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in solving inequalities, how do i know if the solution is all real numbers or no solution?
Just do problems
Really depends on the equation
can u give an ex
well if u know that the only possible solutions is all real or no solution, you can just plug any number into the inequalities and test if it is true
if there are more possible solutions it really depends on the problem
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find min int c
$a^2 + b^2 = c^2 < (a+b)^2$
artemetra
so $a^2 + b^2 < a^2 + 2ab + b^2$\
$2ab > 0$
artemetra
okay this is not very useful haha
(20, 21, 29) works i guess
A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A triangle whose side lengths are a Pythagorean triple is a right triangle and call...
actually (15, 20, 25) works too
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pretty certain this is the smallest
yez
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Is the reason why both of these are equal that
It's already equal to zero
Like because the equation Is already set equal to zero at the start it's considered the answer alr
Something you shouldn't be able to do if they weren't all variables unless you had the answer of what it solved out to
?
Basically are those already the zeroes of the problem
multiplying both sides of the equation by a constant doesn’t affect the equation (obviously can’t multiply by zero but yea)
Why
because either side of the equation is equal to other side consider the following equation: x=2
if i multiply both sides by 2
i get 2x = 4
but notice x was just 2
so 2x = 2(2) which is still the same as the right side
There probably is some hidden goal or representation it has ppl don't explain causing all prior explanations to become annulled.
in your case we are solving for x
when solving a quadratic for the zeros you’re solving for the values of x such that the function equals zero
One second I'm in an airport.
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How do I solve this integral? Everything I tried doesn’t really work
you need to sub back in u = x^2 + 1
cause the end points of the integral are defined in terms of x, you're integrating from x = -infinity to x = infinity
You also need to split up the integral
yeah when you integrate to infinity you should split it up so that each end has a limit
How do you determine that I need to split it up?
when do I split vs don’t split and where do I split it at
If it goes from -inf to inf
Split it up
Using any value c
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**For which value(s) of a does g(x) increase or decrease without bound as x approaches a or for which value(s) of a is g(x) not defined? (Select all that apply.) **
I have no idea what the hell is it is asking. I'm reviewing my notes from class and the professor never talked about this
The question is just asking about the limits as x approaches a
Are you familiar with when limits exist or not
Yes.
A limit exists if it is the same from both sides and not DNE
Oh nvm I see the question asks about g(x) not the limit of g(x)
Didnt you just make a small mistake when entering your answer
I literally am just guessing
You said for a=0?
I have no idea what they want
Aaah
The professor never gave us anything in class about increase/decrease without bound as (x) approaches (a)
that means it has a vertical asymptote there (because it means the limit approaching a is +/- infinity)
So then just -1?
I think it is just x=-1, the entire picture isnt very clear but because it doesnt show an end I think theyre trying to say it goed infinitely further (down)
I tried -1 already:
You forgot second part of the question now
Oh
For which values of a is g(a) not defined
So -1, -2, and 1?
Why 1?
Yes
so just -1 and -2 then
I think so yes
That worked
Now:
From the given graph of g, state the numbers at which g is discontinuous.
Would that be:
-2, -1, 0, 1?
I think that is correct again yes
Got it, thank you
I'll probably be back later, going to close thread until I run into anymore issues
.close
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this is for a part of my problem in physics
how do u solve this to be honest sry
Like do i need the unit circle?
becuz in the video he just got -75 without a second to think
whats Rx?
Show the original question
and the bx was -bcos=-50cos60 i show whole pciture
sorry your handwriting is confusing but thats bx, right?
are you just looking for calculation of bx
this part specific
]
thats the parts before
yes Bx
I just want to know why rx=-75
becuz he did it in 1 second
yw
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@orchid ridge Has your question been resolved?
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Efficiency
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Could someone explain what the answer would be for this?
<@&286206848099549185>
!15m
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@nimble portal Has your question been resolved?
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Yeah you're right
it is the answer
hiii hru
160$ textbook btw
skill issue pirate it instead
who doesnt
anyhow you know how the answer came about right
do you need me to tell you how or you know it?
yeahhh haha im just revising
i forgot everything
in physics
@mossy wigeon Has your question been resolved?
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For twenty the answer is given as 10 but I'm getting 13
Can anyone tell me the way to get 121 in 10 steps?
Nvm got it
.close
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Can i have a hint for the "only if" part?
For the if part, i used the fact that the vertex angles are the same and also the legs are the same in an isosceles trapezoid and so the two triangles with the longest side as the diagonals are congruent
I think that's correct
Diagram:
@exotic cosmos Has your question been resolved?
@exotic cosmos Has your question been resolved?
@exotic cosmos Has your question been resolved?
You can start the if and only if part by assuming that the diagonals are congruent AND the trapezoid is not isosceles
then, to prove the statement, you need to arrive at the conclusion that even if you assumed it was not isosceles, the rest of the conditions make you conclude that it has to be isosceles
@exotic cosmos Has your question been resolved?
yes i'm aware of what only if means
i need help actually proving it
ah using the hint, so triangles DBY and CAX are congruent by RHS (diagonals are the same length, and AX = BY by definition of a trapezoid)
so that must mean that DY = XC or that DX + XY = XY + YC or DX = YC
so then as AX = BY, by Pythagoras we have that DA = CB as desired
Oh thanks, but what's the longer route?
why do you want a longer route
well, how did you approach the question otherwise then
I thought the hint was pretty nice as it seemed impossible otherwise
yeah 😭 but how would you know to drop a perpendicular
the hint tells you
yeah but 😭 idk if there wasn't any hint i would've never considered doing that tbf
if there are right triangles you have to drop a perpendicular
that's why it's there to guide you
mmmm okay, do you have any idea about the "little longer" route?
Maybe it's more natural but with more angle chasing or something
I don't know, you could try opening a thread on AOPS
I considered angle ABD = angle CDB and so on
maybe that's what they are referring to
Oh, i guess it's a bit hard then
Thanks
np
it seemed pretty fun once you realise the hint helps
also there's actually more than one definition of an isosceles trapezoid
Most of the time i do it myself but the hint always guides you to the nicest way to the solution
This one was unapproachable 
Wait something like this works?
Can we show that thing is a square?
But okay that's almost similar to their solution maybe
okay nvm i'll just come back to this when i'm better
Now i'll live with the solution u and them provided

thanks
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Back
Had to get on the airplane last time.
pt1 ANSWER: Because everything including the input (x) is scaled down equally the functions result is still equal to zero even if the function has different outputs given different values.
Y/N
Is this true because the function was already set equal to zero/is zero (they already found the zeroes since its just variables and they can set it to anything they want)
What is the nature of the equal sign and moving stuff past it such as -c/a? Clearly that would be unequal and very strange if it directly meant equal and Ive been given examples in which it appears as though it is indeed not the traditional equivalency sign because
5 + 10 = 15 15-10 = 5
vs
5 + (10 was subtracted/removed) = -10 resulting in -5 once 5 is added. (operation left incomplete and 10 was removed then subtracted)
1st example: the operations are performed normally, everything is equal in the end.
2nd example results in -5 because 10 was removed and then subtracted```
?
?
what exactly is that example supposed to show
5 + 10 = 15 and 5 = 15 - 10
both are true
what's wrong
they arent the same
those are completely different operations
They are statements. Not the same statements, but equivalent statement.
Equivalent means that that if one of them is true, then the other one is true as well.
5 + 10 = 15 is true and so is 5 = 15 - 10. Both are true.
They aren't the same, yes. But is that a problem? Should all statements be same?
you probably really dont try to understand why confused people are confused nor try to resolve from that
not that you arent helping by doing this
i do have another question to ask you
if im being honest for the longest time i already was absolutely infuriated by this because from my perspective it always looks like everyone explains everything in the most complex possible terms and lays out some advanced gigantic block of random thesis and numbers every time someone asks a basic question thinking theyd somehow understand when they couldnt even understand the first step
not that you did it this time
im speaking in general that this happens so often
and what i cant tell is if its actually their fault or not
because every single time ive ever solved a problem it always ends up summing up to the most simple and basic two sentence explanation possible
leading me to see them as almost horrible people when it comes to explaining things
what i want to know is
do you think thats actually the case do they just not care or is it just perspective and they really are doing their best?
This is not a common problem though. You are doubting the foundations here
you are doubting something that most people take as granted
people seem to care so much about whats right they forget entirely about the person theyre actually trying to explain to and that if that person is confused or has even 1 piece missing all information is null because their context itself would be faulted meaning the same information would have a different meaning.
and if you want an in-depth understanding of that, it probably will be explained complexly
because 10 was removed and then subtracted
what do you mean by this
for example you may not of realised a very obvious conclusion that i probably just made a very simple assumption
also i dont ask this because im angry ive been seriously thinking about this for a very long time i just could never answer it.
you ask arbitrary questions, you get arbitrary answers, simple as that
if you're trolling then good job you got your two seconds of fame
otherwise frame your question properly and stop pretending like you're better than everyone else
kheeri i become perfect because im so incredibly flawed that it becomes agony and i have to develop myself to rediculous levels to survive.
i always desire to achieve that perfection whenever i fail in life.
<@&268886789983436800> this discussion is turning unproductive and toxic
i am not better than you
i am a flawed person trying to be the best i can because i cant live with my inability.
not really
i was just asking a question i always wanted to know the answer to.
kheeri misinterpreted it as me thinking i was superior when it was the opposite though its both simutaneously sometimes its a complex topic really.
you did not answer my question
we are i was asking 2 different questions and that one question ive been wondering for like 5 months it makes it painfully hard to pay attention at times not knowing the answer because i always think about it.
to sum up what i mean
they a r e different yes
they dont have to be equal either
what don't have to be equal
the question was meant to see IF they were equal because the context they were given in if it wasnt equal it would cause problems
because thats in context of quadratics where you have to find the zeroes
and if it isnt equal then you dont get the answer
1. quadratics require you to solve for the zeroes so if one value changed you wouldnt have a question equivalent to the zeroes```
this is why i care so much about if it was or wasnt equal
Let's address the 5+10=15 vs 15-10=5 thing first
why exactly do u think these statements are not equivalent
also kheeri stop thinking people are trolling because im not sure if you realise that but you yourself are actually trolling by assuming someone is trolling and disregarding their points and labeling them that when they have a real question
basically
if you have a quadratic equal to 0 and do the exact same operations on both sides (lets just assume for now that this is just adding/subtracting/multiplying/dividing by non zero constants, since stuff like squaring can be weird) then whatever x values (or roots) solved the initial equation, the same ones will solve the equation made after doing some operations
what??
5+10 = 15 vs 5 + = -10
edited
10 IS -10 in this context
thats why it appears nonsensical
how can 10 be -10 C/A = -C/A thats a completely different result
sorry if i miswrote it earlier
5 + = -10 has 0 meaning
well sorry
did you miswrite this aswell? i dont understand it
but thats what they just did
wait
oh so 0/a and -c/a were separate?
to be fair my eyes arent exactly good but wow
i really do have problems huh....
in this case it refers to what i wrote about , first multiplying both sides by a constant a and then adding a constant c to both sides
i wasnt lying when i said i might have dyscalculia LOL
the x values that solve both equations are equivalent
no i just sent the images in the wrong order
im stupid
this is a mess!!
but not really my fault its a simple error
ok so 0/a became -c/a and c/a became negative and past the equal sign
where can i find information on this
$$\begin{aligned}
ax^2+bx+c&=0\
\implies \frac{ax^2+bx+c}{a}&=\frac{0}{a}\ (\text{assuming }a\ne 0)\
\implies \frac{ax^2}{a}+\frac{bx}{a}+\frac{c}{a}&=0\
\implies x^2 +\frac{b}{a}x&=-\frac{c}{a}\end{aligned}$$
kheerii
does this help clear up things
i thought that was your question
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in my lesson I have if (Un) and (Vn) 2 sequences of constant sign from a certain rank and monotone then if Un ~ Vn we have the series of Un and the serie of Vn of the same nature . Can we remove the monotone hypothesis if not do you have a counter example
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hey
anybody understands how we came up to this conclusion?
this regards complex numbers where we're trying to find the solutions of this equation
U^2 = Z where Z is a complex number
suppose Z = a + bi and U = x + yi
,, x^2-y^2 = a \Rightarrow y^2 = x^2-a \text{ then } x^2 + (x^2-a) = \sqrt{a^2+b^2} \Rightarrow 2x^2 = a+\sqrt{a^2+b^2}
bacc
Same if you solve for x^2 instead and plug that in
good goku
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Hello.
Quick question, am I allowed to do this in this specific case?
$\int_{0}^{1} \sum_{n=1}^{\infty} \frac{(-1)^{2n} \cdot n^{-1+n} \cdot x^{-1+n} \cdot (\ln x)^{-1+n}}{n!} , dx$
Mary
$\sum_{n=1}^{\infty} \int_{0}^{1} \frac{(-1)^{2n} \cdot n^{-1+n} \cdot x^{-1+n} \cdot (\ln x)^{-1+n}}{n!} , dx$
Mary
I am assuming I can, just wanted to double check to make sure.
(nx log(x))^n / n! convergence looks sus
i wouldn't "assume" you can
That was my only worry
like it's not immediately obvious you can
do you have the original problem statement
This works comes from here
$\int_{0}^{1} \left(x^x\right)^{\left(x^x\right)^{\left(x^x\right)^{\cdots}}} , dx$
Mary
I would definitely say it does
@hazy trout Has your question been resolved?
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do you know how to plot linear and quadratic equations?
Uhhh
No
Well I probably do
But I don’t remember how to
Once you explain it may come back to me
@wraith robin
What do you get when x = 1
okay maybe go on khan academy and learn how to plot an equation first, they have good resources
You get a straight vertical line at x=1
Okay
check again
that is plotting the line x = 1, not the value of the function at x=1, which i believe was the question being asked
Oh
best bet is to go through khan academy algebra 1, itll actually teach how to read these equations, what they mean, and how to visualise them
Okay, would it be okay if I ping you once I’ve gone through that?
So just go through all this?
yeah j go through the resources on that website and you will be able to answer the q urself easily
yeah i was asking what g(1) is not what a graph y = 1 looks like
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How do I know what the inside and outside functions are for chain rule and I’m a little lost on what to do in general
could you elaborate on how you're confused? The way I remember is that inner functions are harder to reach
So the inner function would just be -ax?
yeah you're right
I was confused because it was an exponent and I wasn’t rlly sure how that worked
you turn it into f(u) = ae^u and u = -ax and continue from there.
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Why is the angle between u and w 120degrees while the angle between u and v is 60degrees?
for # 11
Because you have to rotate u 120 degrees to turn it into w
wouldn't you have to flip it 180 degrees?
No
when you're looking for inner multi of two vectors, make sure the two vectors start from the same spot
and don't forget vectors can be shifted
ah ok
A vector doesn't have a start point and an end point
It just has a direction and a distance
but still, in order to make sure the two vectors start from the same point, wouldn't we have to rotate the vector 180 degrees?
No
You would move it
If you rotate it it's no longer the same vector
If you move it, it is
Because vectors don't actually have start and end points
But they have directions
I see... so I would just shift u down to meet with the starting point of w
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Np
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Why was 0^0 evaluated as 1 here?
It's just convention
But one reason could be because the limit of x^x as x goes to zero from the right is 1
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Shown is the graph of a function f(x). Sketch the graph of f′(x) by estimating the derivative
at a number of points in the interval: estimate the derivative at regular intervals from one
end of the interval to the other, and also at “special” points, as when the derivative is zero.
Make sure you indicate any places where the derivative does not exist.
@burnt gazelle Has your question been resolved?
<@&286206848099549185>
this?
yeah
wow, this question is really really cool. I however dont know how to answer it. so like derrivatives are like guessdimation based on how far you wanna get into it all. so like at -0.8, the function is 0.8. and like at -0.6 the number is 1.2. at -0.4 the number is 1.3 and so on. these might be special points. I dont know if any of what I said was helpful. I dont know too much of how this stuff works.
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Approximate a root of f = x^4 + x^3 − 5x + 1 to one decimal place.
First of all, this is from a book and I already know on of the answers(which is .2), but I want to get to that answer by myself, I tried to use Newton Raphson, but there are too many decimals and in the test I can not use a calculator
Our teacher said that we have to do it by guessing
Sure, for example I chose [0, 1]
But ho do you get to that .2\
Did I explain myself?
well you can round along the way
nothing
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the question is to prove/ disprove the statement above in the rectangle,
but when I saw this disproof i got confused.
why does polar coordinates not work here?
(the limit in polar =0 but the original limit does Not exist)
can anyone please help me with this or tell me when do polar coordinates work and when not?
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@gleaming panther Has your question been resolved?
I think that phi is not fixed here, this is why
The assumption is only for phi fixed
but why does the polar coordinates method not work on this function?
like why even though it =0 the original limit is not 0 and does not exist
You are not considering the same paths
Having the angle fixed means that you are doing limit on straight lines
Here, you don't
You just consider other functions in a sens
f°gamma, where gamma is either a straight line or gamma(t) = (t, k cbrt(t) )
but like I mean why does polar work on this function but does not for the func' from before?
like what is special about that func' that polar doesnt work
or is it like because the limit is not F(r) * G(theta) and G(theta) is bounded?
I would say that, yes
Or it touches with that
ok ty
i got confused cuz my lecturer did not go deeply into this whole polar thing
tysm
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∆ABC has vertices A(3,4), B(−5,2), and C(1, −4). Determine an equation for
a) the median from C to AB
On what exactly are you stuck?
I need to make a midpoint then make a slope out of the median right?]
its just my answers dont match to the questions answers
Oh give me a second
I got this
how did u get it
So median is a linear function
And has to have a form $y = ax+b$ for any point $(x,y)$ on it
Goran
Meaning that you can get two equations to find a and b as: $3 = -a+b$ and $-4 = a + b$ which you can pretty easily solve
Goran
Find midpoint of AB
so I need to ask you to like
say it simpler
midpoint bro
ok
(-2,3)
tru?
The midpoint should be (-1,3)
Maybe that's why your solution is wrong
A and B midpoint no?
Yeah
I did C and A
Yeah
Ahh
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Am I crazy or does this problem not make sense? how can it be aroudn the y-axis when deriving in terms of x? furthermore, if it was a typo and it's around the x-axis, then why is this not the integrand?
this is the shell method i believe, there is no typo
oh the surface, didn't catch that
still, shouldn't you be able to find the surface area with a method analogous to the shell method ?
Mayhaps, but this is the formula used in this chapter for which the homework was assigned:
S = ∫[a,b] 2πf(x)√(1 + (f'(x))^2) dx
on second thought no it's impossible using shell method
i dont see why it would be
considering cylinders forming around the axis with a crossection similar to that rectangle drawn, I think no
idk how to explain why I don't think so but that's definitely not what it wants
why cylinders ? you're not finding volume
ya
okay this is a tangent lmao
i mean the integral is just going to be finding the surface area of a section of a cone
it wants me to use
$$S = 2\pi \int_{a}^{b} f(x) \sqrt{1+(f'(x))^2} , dx $$
Dominick
For example, this problem:

