#help-17
1 messages · Page 191 of 1
Okay here is how far I've gotten without not knowing what to do next.
Can you see it?
yes
it looks like you just copied down the vector field given 😵
let's start from here
I'm just breaking down the vector field $\vec F$ into $L \ dx + M \ dy$
Ari
Since that's the setup for Green's Theorem
And then I evaluate these partial derivatives to get what's on the right hand side under the double integral
Okay thanks! I'll check if it works in a few minutes!
What is dx and dy? Do we solve in relation to t?
no
we're just separating the vector field $\mathbf{F}(x,y) = (e^{-x}+y^2, e^{-y}+x^2)$.
Ari
the left one is dx, the right one is dy
did you draw the curve
that'll help you see the bounds of the double integral
So for dy it is probably 0 to cos(x) and for dx it is probably -pi/2 to pi/2 right?
Am I right?
correct
Okay thanks!
the integral is a bit of a pain to solve but it evaluates nicely in the end

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help
my original thought would be to say that let (xn) be some sequence of an in (an). Then (xn) is either a permutation, subsequence and thus must converge to a
someone told me that was wrong and to split it into two cases: when image of (xn) is infinite and when it's finite
when it's finite, assume the limit of xn isn't in the set and then say something about two epsilon balls
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This kind of question can be solved by factoring right? You only use the quadratic formula when there is an equal sign present ( someone pls confirm this)
you can factor yes
the quadratic formula is used to find roots of a quadratic polynomial, but it can technically be used here too
you can use quad formula
there's nothing wrong with using it
find critical values and do a numberline test
for this particular problem though, you can just factor
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🙏 I’ve lowk been asking the same question but I js wanna make sure: for question 1, using the quadratic formula would be easier bc the = makes it easier to verify by having a variable on both sides. But you could still use the X factor method to find x values
For question 2, factoring using the x method would be easier, but there’s nothing wrong with using the quadratic formula
Let me know if my question is confusing 🙏
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bro that's $$ (x+3).(x-5)<= 0
which ultimately led to (x+3).<=0 and (x-5) <=0.. now it seems to be solved
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I am getting the wrong answer for (b)
sending my working in 1 min
the correct answer is 48 but i am getting 62 over and over again
bro that 2a is in addition on the RHS how can you cancel 2s by that 2
did you get it ?
@finite trout Has your question been resolved?
just for simplification... don't move the variables at first.. just put values in first eqn then after try to solve
and find value for a
if you didn't get it show me your work.
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If a man had walked 2 km/h faster , he would have taken 40 minutes less to walk 5 km.
find the walking speed.
but would that really matter
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yw
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the answer is C
but i have no idea how
its just logic btw, x:y dont have any special meaning i think.
420 
oh yea you're right
thanks
also struggling to figure out the logic in this one
I tried to check number of letters between each pair, but i dont see any pattern
ah i got the answer
i kinda googled it
pretty odd tho, wouldnt have gotten it on my own
anyways solution if anyone was curious
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$$\int \frac{cos(\sqrt{t})}{\sqrt{t}} $$
Nerdy_Coder
Nerdy_Coder
Nerdy_Coder
I know how to solve the Fibonacci sequence!
Im guessing your derivative isn’t right
du = 1/2*t^(-1/2) dt
2du = t^(-1/2) dt
My uhh book said to solve it like that
Uhh...hold on...
Why 1/2t
no makes sense
the u = sqrt(t) is right
and then you derive it using chain rule
What's dx
iuck
???
how?
u=sqrt(t)
du/dt = 1/2sqrt(t)
dt=2sqrt(t)du=2udu
change dt to du
you get 2ucos(u)/u
=2cos(u)
Let him answer
1/2 t^(-1/2)
$\int cos(u) du = \int cos(u) \frac{du}{dt}dt$
iuck
And u = √t here
Warning though you can treat du/dt as a fraction but technically it isn't
It's not "cancelling the dt" here to be specific
Where did the other u go?
What u?
I worked backwards
Just put u=√t in RHS
And you'll get the question
du/dt is 1/(2√t)
Basically if I substitute u=√t, I'll get the LHS
And as a result you need to calculate du/dt
To perform the u sub
Why does calc need intuition the thing is for calc at least I lack intuition for it completely
Of course there are some parts that may not feel intuitive
But practicing more problems will help
Like for example at this stage one cannot show (du/dt)dt = du (requires advanced math)
We just do this in u sub
..uhh what?
I'm confused
I have been practicing and uhh couldn't do this problem
Let's do a simpler one then
Will this build the intuition expected of me by my textbook?
The stewart one?
I never did Stewart sorry
Then I just failed badly
Happens
I have lost my confidence to teach myself math but I have too
I suck at it and need to learn it for my degree
You can see the RHS as $2\int f'(t)cos(f(t))dt$ if it helps
iuck
It doesn't
So at what part are you getting confused
u substition applying to a non-integral
"non-integral"?
Well nothing actually lose of confidence to learning math
I tried practice and uhh not going anywhere
idk what to do
but I need to study
practice makes perfect...is clearly not working
Take a walk outside
The "perfection" is a long term result
Not a short term one
I suck at this
I did as well
Should I just give up on learning from my textbook?
or just ask for help on any problems I cannot do?
Not give up for sure, but you can try other textbooks as well
Of course you can ask for help here
I understand the concepts at a basic level at least
textbook is best resource most times. Doing exercises is next step. Asking questions is next step after that. Repeat
That's good, don't get anxious
So math is something learnt in the classroom and with other students/tutors?
As with coding I can just pick up a book read it go thru it read the reference gitgud
There's a different between "knowing" and "understanding". Based on your description, you "know" them but may not yet "understand" them (not to get philsophical on you)
It's best when done like that, of course you can self study math as well
...math I tried that only works for basic algebra
Well i tried and uhh clearly I am not capable of it?
For most people, yes. That's all learning in general
Are there those that can?
I self study
Well no i can teach myself things.
But I love math
Try certain textbooks made specifically for self study
How far did you go?
What is stewart made for?
as I said, I never did Stewart sorry
Nor did I self study calculus as well
Did you teach yourself real analysis?
Because I didn't had time due to my entrance exams
I didn't even do real analysis yet
I have some self study there too. But I never finished it. I know enough to read about topology, and I fgo back and forth whenever I get bored
Don't pressurize yourself that much, there is time pressure for sure but don't get too anxious
If you want, for now skip this problem and move on to the next one
...Should I do that?
I keep getting stuck and losing all confidence
Should go back and study algebra?
I uhh forget analytic geometry
and uhh made mistakes with equations all the time
Your wish
I lost all confidence in myself to learn math
Is this jsut an excuse?
To do what?
Guess have fun and but I want to study
uhh
....
When I get stuck do I just go to someone else for help
because videos and other resources have stopped working?
It's about understanding the concept. The book/resources can excellently show the concept and its working, but it also depends upon the person as well to understand it
Sometimes the person can have a brainfog/get tired to digest more concepts
I don't get it then
That's why at times like this I take a small break, or skip the problem to revisit them at the end of the day
So...just ask for help right?
Because it seems I am no longer able to self-study math
as...I have hit the wall that people in other fields say people who self-teach hit...
Try the problem a few times yourself then ask. Try all the possible methods you have learned before asking
Wait maybe that's my answer?
Yea I should stop trying to just teach myself
I should be asking for help all the time until i can self-stufy again and repeat...
I don't like this idea but this is likely the only way
Try to join a study group as well if possible and if it works out for you
...hmm well my college does have a tutoring system
I should set up 2 meetings every week if possible
right?
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does anyone know how to graph this in one equation?
probably $-20e^{\frac{-x^2}{2}}$
ƒ(Why am. I here)=I don't Know
,w graph $-20e^{\frac{-x^2}{2}}$
Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online
then $ne^\frac{-x^2}{a}$
ƒ(Why am. I here)=I don't Know
how could i make this one function
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how to make it independent from y
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hello! can someone help me w this?
Can you plug in g(x) into the given f(x)?
but idk what g(x) is
Let g(x) = t, you could get f(t)
Yeah, so just plug in g(x)
As it is
$f(g(x)) = g(x)^2 - 4 \overset ! = 4x^2 - 8x$
Right?
Kepe
Now solve for g(x)
thanks!
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Got this question here asking about critical points. The options are : there are no critical points and the others are what kind of critical point is (0, 0). My intuition is it is a saddle point, but I'm not sure of the best way to verify my intuition.
I don't need to get to deep into analytical methods since it is a MCQ
You will need to first find grad f.
Then you will need determine when grad f is the zero vector
Those are done, it is the point (0,0)
Followed by that you need to know the theorem related to the hessian matrix
Ok, I was trying to avoid doing using the hessian matrix
I thought there might be some trick to infer it quickly
maybe
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(A) will arrive in a restaurant in X time and (B) will arrive in Z time. The joint probability density function of 𝑋 and 𝑍 is constant over the set of values 𝑆. Determine the joint probability density function of 𝑋 and 𝑍 and find the marginal probability density functions of 𝑋 and 𝑍.
im not a helper but 1 question why is the integral from 0 to x?
bro really i dont understand this, in my course this is how in the lectures i have seen this
it doesnt work like that
the fz(z) looks fine to you?
nope 😭
how would you do it
is fz(z)=integral from 0 to 2 of 1/2?
yes
what do u mean
thats all i needed to hear lol
how can i find fx(x) and fz(z)?
first of all its called f(x) only thats it
what grade are you?
uni first year
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I try to understand Universal Approximation Theorem but before understanding the formal proof, I would to understand this:
https://www.desmos.com/calculator/cfvtjusqmq?lang=de
how can it be that adding n_i together approximate the function f ?
Is there a geometrical interpretation or a better intuition?
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can someone explain why when calculating pi you need to make sure the following statement needs to be true:
x^2 + y^2 <= 1.0
The set of points for which x²+y²≤1 is the set of points inside of the circle
Point (x, y) is inside of a circle (centered at origin) iff the distance from origin to the point (x, y) is less than or equal to the radius (in this case 1). The picture below shows that the formula for distance of point (x,y) from origin is sqrt(x^2 + y^2) (pythagorean theorem was applied there). So we are looking for points where distance from origin <= 1, that is sqrt(x^2 + y^2) <= 1. But if we square both sides we get just x^2 + y^2 <= 1, which is the condition
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what does it mean by is a perfect square
like does the formula equal like a square??
if a quadratic is a perfect square, that means it has a double root
a perfect square is a positive number that is the square of another number
for example, 4 is a perfect square because 2^2 = 4
and 64 is a perfect square because 8^2 = 64
ohh
hello Moosey
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you can also think of it as ax^2+bx+c=(x+k)^2
thanks
I was gonna bring this up 

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I'm doing some exercise with interest and stumbled with this problem. I did two ways to solve it which is a shortcut i just found out which is the correct answer and a traditional one. For the shortcut one i just simply divide the non zero and then i will move the decimal place of the answer(0.6) based on how many zero is left so I got 0.06. On my second attempt, I added 2 zeros because you can't divide 3 with 50 and then I got the same 0.6 as answer and I think I messed up with decimal places on my 2nd attempt.
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Can someone please show me how to do this, the function is “y = x^2 + x - 6”
I’m mainly struggling with finding the difference between task 2 and task 3 and also the fact that drawing the secants is hard because of how steep the line is
You can see here that i drew it but you can barely even tell the difference betwen the secant and the actual graph
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I do the partial derivatives, and get the critical point (1,0)
and solve for that, which gives me -1
then I am lost
you now need to look at f on the boundary of D
and find the min/max from the boundary points
then compare with the critical points you got on the inside
meaning
extremas are either on the inside or on the contour of D
well you have a few options
for example, you could express f(x,y) as f(x) when (x,y) is on the segment between points (2,0) and (0,2)
so any point between (2,0) and (0,2)
and find the abs minimum of "f(x)" on [0,2]
any of those points that lies on the LINE formed by those points
(so just find equation of a line)
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is 1/2 greater, smaller or equal to -4/x
i just need the answer
the question is like saying use < or > or = and like assume that each variable is positive
so i would assume that x could be 1
so issit < or > or =
yes 
wdym
xd
sorry sir, we don't just give you the answer 
if you think about this, it might lead you to the answer though
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In R^n, if a set S is composed of n linearly independent vectors, does that mean S spans R^n?
Yes
damnit
In general for finite-dimensional vector spaces, any linearly independent set with the number of vectors equal to the dimension is a spanning set
Hi, if you're satisfied by the answer, please close it with .close. Thanks.
this can be proven using the steinitz exchange lemma if you want to know more about
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Can someone help me with this
Do you know of any sum or difference identities regarding ANY of the inverse trig functions?
Guys, this is an acute isosceles right?
!occupied
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How do I solve the 3rd question🙏
Is that Gaussian symbol? Or just bracket
Mind if I ask
Yes sire
Why the result is not an integer
Oh yeah mb
The answer
Is null
This one actually wait
The fifth one
$$[x/2 + 5/3] = 2$$
SusGusInaBus
Hoho there
whats the brackets for ?
Gaussian notation
Greatest Integer func
No problem
You mean it rounds up?
Only keeps the integer
Oh so rounds down
So you know the inside is between [2,3)
2 included, 3 not included
Try to make an inequality
Oh k
Please #❓how-to-get-help
and what is a .heic file
can you take a screenshot
yea
.close
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Where did you get it from?
old exams from my university
And does it want a numerical value?
no it asks me to check if it exists
Oh okay
(converges or diverges ig)
he has many simpler problems and i have no idea how to aproach them
Can you find the derivative of that expression?
the derivative yea i guess i could with some time
but how would that be useful?
this type of question is always the last question on his exams, must be the "super hard" one
So that we can visualise it
Maybe a comparison test could work though now that I think of it
such as?
The top is always between 2 and 3
Set it to 3, which will give a larger area/integral
If that integral is convergent, so is this one
oh yea that could work
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It is a multivariable chain rule question
I'm just starting with this, it came out very complicated, so I wanted to check what is the best way to tackle it, if there are any techniques or details I am missing
As a more concrete question: when I reach the end of the chain rule I've got an expression with x,y,z and r,st. Do I need to change everything to r,s,t or can I just evaluate x,y,z at the point and substitute that?
Here is my solution
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are these rules true? If yes do they have a name?
these are a consequence of logarithms being the inverse operation of exponentiation
MethIsAlwaysRight
. instead of !
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i intergrated and got $\frac{1}{2}ln(b^2+4)-\frac{1}{2}ln(4)=ln(\sqrt{5})$ then $4^{0.5}$ is $2$ so i just wrote it as $ln(2)$ so $ln(\sqrt{5})+ln(2)=\frac{1}{2}ln(b^2+4)$ then i did $ln(2\sqrt{5}) = ln((b^2+4)^{\frac{1}{2}})$ then $ 2\sqrt{5} = \sqrt{b^2+4}$ then 2\sqrt{5}=b+2$ and finally $2\sqrt{5}-2=b$. where did i go wrong?
morphine_addiction
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sqrt(b^2 + 4) is not necessarily equal to b + 2
+-b +-2?
Still no
what is it?
It's itself, you can't simplify it further
should i square 2srt(5) instead?
Yes
Yes
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I need help I dont understand how to factor at all in relation to algebraic formulas
Post an example
6x^4-7x^2+2=0
I understand that u would take its place
I get all the way to 6u^2-7u+2= 0 and dont understand anything about factoring
or how to factor a problem
12
then 3 and 4 give you 7
see I have tried this but for some reason and I dont know why but my math always ends up wrong
and how do you know what coefficient in front of the (u-3) (u-4) is supposed to be
faiyrose
how do you factor by grouping?
(3u-3) (3u-1)
for the first term
yes
the ^2
yes
idk
how cna you tell if it can be factored?
I'm sorry I'm really confused
why is it in a fraction now and why do we have some many 3u's
yes
is the ? the value of u?
or x
ok
isnt it just 3u
oh
no
cause 3 x 3 is 9
so its 2u
is 0
well 1
faiyrose
yes
faiyrose
one second I am writing this down
ok ready
is it going to be -2u(2u+1)
oh
-2u(2u-1)
wait but how would you get 4u+2 with u(2u+1)?
oh I see
cause
that would be 4u^2
so it would be 2(2u-1)
oh yeah
cause you want a positive
2
and negative 4u
one second
so how are you factoring out only 1 of the 2u-1's
ok because its the product of us factoring 6x^2
-3u
what's 2u*3u?
6u^2
yep
I haven't done it yet because I dont understand where the (3u-2) came from
(2u-1)(3u)-(2u-1)(2) =(2u-1)(3u-2)
shouldn't it look like that?
ok I am following now why does that make (2u-1)(3u-2)
I dont thinks so right?
4x^4-1
2x^2
3x^2-2
what is the zero product property?
ok so how do I solve for x with that ?
that just means that x=0
faiyrose
they want it in a solution set
so (+- the root of 1/2 , +- the root of 2/3) ?
i2/3, i1/2
faiyrose
faiyrose
where just saying the product was (+-the root of 1/2, +- the root of 2/3)
yes I dont understand how you got this from 1/2 and 2/3
what does that mean
why is the denominator irrational ?
yes why is the 2 the only thing being squared?
faiyrose
faiyrose
1
faiyrose
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Q
What’s a translation in terms of transformation
for a function y = f(x) then a vertical translation "b" units up is y = f(x) + b and a horizontal translation "a" units right is y = f(x - a)
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Hello
Hello
hi
I’m an expert of high school trigonometry
Today I encountered a problem that seems to be beyond my ability to solve
I’m astonished
I’m endowed with the ability to graph any given trig function in my head within a second
But
I failed on this one
Do you guys agree that the period is 20
It should be deemed as a violation against the wisdom of humanity to neglect the question from a supreme trigonometry master
Oh it is solved
It is indeed 20 btw
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What’s up
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hii guys
\begin{align*}
\pi \int_{-1}^{3} [8 + 2y - y^2]^2 \, dy &= \pi \int_{-1}^{3} [64 + 32y - 4y^2 - 4y^3 + y^4] \, dy \\
&= \pi \left( \int_{-1}^{3} 64 \, dy + \int_{-1}^{3} 32y \, dy - \int_{-1}^{3} 4y^2 \, dy - \int_{-1}^{3} 4y^3 \, dy + \int_{-1}^{3} y^4 \, dy \right) \\
&= \pi \left( 64y + 16y^2 - \frac{4y^3}{3} - y^4 + \frac{y^5}{5} \Bigg|_{-1}^{3} \right) \\
&= \pi \left[ (192 + 144 - 36 - 81 + \frac{243}{5})- (-64 + 16 - \frac{4}{3} -1 -\frac{1}{5})\right] \\
&= \pi \left[ (267.6) - (-50.53) \right] \\
&= \pi (318.13) \\
&= 999.435
\end{align*}```
Juan
im doing something wrong dont know why, im sure until the y replacement im right
but dont know after
the asnwer should be
,w \pi \int_{-1}^{3} [8 + 2y - y^2]^2
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could you please help me understand"forming algebraic expression
And teach me how to do 1 and to 2 step Operation
Post the question
yes and wht you have tried so far
@safe tundra Has your question been resolved?
do you want to start with no 8
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I know finance is a sensitive topic and there are always strong opinions about it but I just really want to keep it simple to the logic i'm hoping to achieve if it's possible. Just been struggling with it for a while now and I know people who have a stronger background in math can do provide some good perspective. I want to keep it brief and as simple as I can but I need to give a background to hopefully give some perspective on what i'm hoping to achieve.
I'm trying to create what they call a "dynamic trading range" that can be represented by supports/resistances. A support is basically an area where the price generally seems attractive to buy at, so a lot of people will be buying a stock at that price which will consequently prevent a price from going lower. A resistance is the opposite where people would be more inclined to sell. I do believe that the support/resistance areas are of higher predictability (lower entropy) in general so I am trying to run a test to prove my hypothesis based on real data. In order to do this, I need a way to automatically map out these areas and the logic to do this is what I need guidance on. The term dynamic is because the trading range is updated every day as soon as new data is added.
The current data which I only have and will ever have access to are the daily stock quotes which generally describe the average behavior of the stock during that day. They are represented by Open (what the price was when the market opened), High (maximum price made within that day), Low (lowest price made within that day), Close (what price the market closed at), and Volume (total number of shares made on that day) for each stock/security. The nature of the data set is already not ideal because making an analysis on these is similar to the concept of taking an average of an average so the conclusions that will be made from these will be suboptimal at best. But it’s better than nothing.
The trading range will also be a function of price, volume, and time. Since it is hard to capture the relevance of the price data due to the nature of the data as mentioned earlier (Open, Low, High, Close) most people just use the either the “Close” price or what they call the typical price which is (Low, High, Close) divided by 3.
The volume is an essential component when defining a trading range because it provides context to price movements and helps validate the strength of the range. An example can be a math class with 100 students and if 95 students scored 33 points, and only 5 students scored 99 points then the average would only be 36.3 points. The price can be represented by the student’s points and volume can be represented by the number of students.
There can be an infinite number of ways to set up the beginning point and end point of the trading range. So this ramps up the complexity of the model a lot because there has to be some way for the model to decide for itself which among all the possible combinations out there are the relevant ones.
This example shows one that only use mininum and maximum values, but this is oversimplified and is of no use at all. I asked AI models a hundred times to iterate it cannot seem to come up with something robust (with respect to the infinite number of zooms). Trying to determine how to mathematically come up with the rule/criteria
These are just examples of where a support and resistance may be at based on today's latest data for a single stock
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Have you tried the sine rule?
Idk how to do it
I just need help in this
Please
Apply A/ sin(A) = B/ sin(B)
Is this from an exam
yeah
sorry, we can't help you in that case
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🤣
Real
bro tried to use obliviate
stop trying to dm me
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Have you learnt it?
Sure
And now im going to fail
@finite fern Has your question been resolved?
@finite fern Hey, if you’re here to get the answer to that question directly, I’m afraid that no one would help you. You may either sit down and answer my question or just leave since you’re not going to get anything good for waiting here:)
bro what do u want from me
Nothing, I’m just notifying you that you’re wasting your time waiting for the answer
That’s it, it’s up to you
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can i get some help on sloving this, i am completely stuck on this
Do you know the Pythagorean theorem ?
yes
Apply it here
so 6 is adjacent?
We can't say it directly
It depends upon the angle which you are taking
But that doesnt matter here
6 and X are the base and height or opposite
a^2+b^2=c^2, plug the corresponding values in and solve
would there be like a formula like $6^2+x^2=3x-1^2$?
Bahnies
Yes (3x-1) ^2
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you're welcome
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i know that 6561 can be factorized as 3^8, but is there a easier trick to it?
given that it's an 8th root you could just start taking small numbers and raising them to the power of 8, if that's any easier than just staring at 6561 and trying to factorise it
there isn't really going to be anything that doesn't go through somehow concluding 6561 = 3^8 since that is essentially what they're asking you to determine
so there is not a easy trick to do the factorization?
well given that you're expecting the answer to be an integer, you can notice that 6561 is a multiple of 3, and then just keep dividing it by 3 to see if it's 3^8
it would be pretty weird if they asked you to find, like, (3^7*5)^0.125, because then the answer is going to be some weird irrational number, and there isn't much you can do about that without a calculator
how do u know that the base is 3?
well you just start trying numbers until you find a factor of 6561
it's not even, so 2 doesn't work
it doesn't end in 0 or 5, so 5 doesn't work
there's a fairly easy trick to check if a number is divisible by 3: add up its digits, and check if the result is a multiple of 3
6 + 5 + 6 + 1 = 18, which is a multiple of 3, so 6561 also is
well if you get numbers that big then i'd say, start computing 8th powers of things just to get a sense of how large it is
like if you want to find the 8th root of 208827064576
10^8 is too small, so it's bigger than 10
20^8 is still too small
but 40^8 is bigger so it's somewhere in between those
30^8 is bigger
etc.
and eventually you'll try something that happens to be the right answer
or actually in this case
the first thing you should do is notice that it's even
and divide it by 2^8, just to make it a smaller number
since 815730721 is at least somewhat easier to deal with
but also yeah i would not expect numbers this big to come up
i am not a calculator lmao
oh okay
well you can divide it by 2 eight times
so i dont need to worry?
yeah i don't think you need to worry
i mean you might be asked to find square roots of things like 23^2 because that's actually a pretty reasonable number, it's 529
but also like, that's just way easier because the numbers are smaller
20^2 is 400 which is too small, 25^2 is 625 which is too big, we can skip the even numbers, 21^2 is 441 which isn't right, so it must be 23^2 and indeed it is
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need help
If you are out of ideas, use the definition of
$x^a = exp(a ln(x))$
Daddy_314
bro tf does the rhs mean
Express tan(27) and tan(39) in terms of d
If you did not study the exponential nor the logarithm function, forget it
yeah im still in eight
what do you mean? if i’m trying to find d, don’t i need the value of the opposite or adjacent?
"need" is a strong word
You "can"
Do whatever you want
If you have a good approach
@kind light Has your question been resolved?
what does it mean to “express” tan(27) and tan(39)?
The red triangle is a right triangle
Therefore tan(39) = h/d
Tangent is the best choice because we have the adjacent side to the angle and we want an information about the opposing side
There is no point in using cos or sin
The second step is to do the same thing we did in another right triangle
so tan(27) = h/25+d??
alright
And now the magic happens because we have two equations and two unknowns
This is a system of linear equations to solve
tan(39°) = h/d
tan(27°) = h/(25+d)
Tan(39) and tan(27) are just two numbers
Like any other number
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(Regarding my electricity bill) Q: What is the period of this bill and how many days of usage did it cover?
hi im kinda struggling with these word problems, ive figured out the first part of the question
billing period: 11 jan - 9 april
is the second part asking how many days are in between those dates or is it asking about some other info
im pretty sure thats what its asking
no problem angela
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hello I am new. so is there a help that helps with precalc or is any help fine?
You can post anything you want.
There are also more topic specific channels below, you could ask there as well. Whatever works for you.
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Want us to grade it
Nothing I’m gonna do it this pic was b4
I’ll send here aswell give me 5 mins
I did the other question I’m done don’t need help on that one it’s fine jus lmk abt these questions if they seem correct
I have to take shower and get ready for math class brb
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