#help-23
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can you show your work?
I don’t really know how to start the problem
$$\int_a^b f(x) \dd x = -\int_b^a f(x) \dd x$$
$$\int_a^c f(x) \dd x + \int_c^b f(x) \dd x = \int_a^b f(x) \dd x$$
NEONPerseus
Do you know these properties
Yeah
Have you tried using them
I just don’t understand how you would get 15
NEONPerseus
NEONPerseus
Using the two given integrals can you find F(8) and F(1)?
I’m blanking
Hmm
If I say the integral of f(x) is F(x), what would the first integral evaluate to
i think there's an easier way to visualize this than with F(x)s

(it's the same thing but keeping it as an integral)
You can do it with these properties I guess
Try to get the bounds of the two integrals in the way that second property is laid out
yeah i think that's enough
Wouldn’t integral 8 to 1 of f(x) just equal 1 tho
how did you get that
After combining the two integrals
ye
Also do you mean 1 to 8
so we have $\int_1^8 f(x), dx = 1$
maximo
Yeah
Just the 2 bit
But wouldn’t your final answer be 3 because you just add them or am I in the complete wrong direction
well
You need to make sure the bounds match up
What you did was correct by flipping the bounds on the first I assume
You're integrating from 1 to 5 and adding on the area from 5 to 8
are you saying $\int_1^8(f(x) +2), dx = 3$?
maximo
Oh
Yea
Nah I think it's good you came in 
have you heard of this $$\int(f(x) + g(x)), dx = \int f(x), dx + \int g(x), dx$$?
maximo
it's part of the linearity of the integral
Yeah
we should apply this here
I still just don’t know where 15 comes from at all
maximo
let's split it like i split this integral
by limit you mean integral yeah?
Yeah
if so yes, it's not just the number (usually)
I’m reviewing all the units right now
They’re mixing together
Thank you for the help
no problem
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I need help with number 9
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
start by finding the inverse of h
how did you get h^-1(x) = 1/2?
Am I wrong?
it would be..
x + 1 = 2y
(x + 1)/2 = y
Sorry my mistake for doing it carelessly
@icy lance
Sorry for pinging you alot..
its alright
yeah h^-1(x)=(x+1)/2
now just use the function as per the question and then solve for m
Oh so would it be like this..
h^-1(5m - 1) - h^-1 (9 + 1)/2 =15?
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Hello
I have a query regarding Pythagoras' theorem
Just ask
L spelling tho
PQRS is a rectangle in which PQ= 9 cm and PS=6cm. T is a point on PQ such that PT= 7cm and RV is the perpendicular from R to ST. Calculate ST and RV.
Notice PTS is a right triangle
same for SVR
SVR is a bit tricky. It'll require construction
or you can get angle VSR and use trig to find the side RV
That'll require calculator
tanx=7/6
@sour hornet Has your question been resolved?
@sour hornet You there?
Yes I am
Yeah so you need to connect R and T
Ok... and then?
Use pythagoreas theorem to find RT
from triangle TQR
and yeah find ST also using pythagoreas
Listen... If you look closely, the triangle TQR is a triangle but I will need at least 2 sides to apply the Pythagoras' theorem
This is the diagram that is supposed to form according to the question and told by my teacher
yeah so you have qr and tq
qr is 6 and tq is 2
Well the answer isn't over
we need to find vr and we have only found tr
Now vr is perpendicular to st
so area of triangle str can be given by 1/2xSTxVR
But area of triangle str is just area of rectangle- area of pts- area of tqr
so equate them
assuming vr=x
What are you labelling as x here?
ok got it
Yeah good
thanks
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No idea where to start with this question, can someone walk me through the process?
nvm i figured it out
i was simplifying my partial sums wrong
i found the pattern
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dumb but would sin6theta= 2 sin3theta cos3theta
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I need a little clarification to help my understanding of the last sentence.
secx = 1/cosx. So if secx = 1 does that mean then that I'm just looking for where cosx = 1?
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I thought tan(0) was undefined at the values where sec=0 which is pi/2 and 3pi/2
zesty&festyII — Today at 2:34 PM
I thought tan(0) was undefined at the values where sec=0 which is pi/2 and 3pi/2
Yep.
So, this is the check against a larger problem given
Where we get x=o
so secx = 1 at (1,0)
And the screen shot above is checking to remove erroneous solutions.
It's saying to plug in 0 into the Pyth ID that was used earlier in the problem.
sec(0) is showing in the check as =1
that still made no sense and isn't really relevant to the question
,W sec(0)
tan(0) is defined, it's value is 0
sec0 is 1/cos right, so wouldn't I be looking at where sec is 0?
sec(x) = 1/cos(x)
so what?
I'm explaining my thought process, which is obviously wrong somewhere.
tan0 is undefined at those values.
(0,1) and (0,-1)
tan(0) is defined, it's value is 0
you don't really care about where cos(x) is 0 here
tangent of (0,1) is sin/cos = 1/0 = undefined
you don't have those values in your potential solutions
Ok, I think my confusion is coming from knowing I need to look at sec0, which I have to transform to 1/cos0.
you can express that in standard trig functions when evaluating/solving you find that more convenient
Yes, agreed. But they're in the check I"m asking about.
overcomplicating/overthinking
tan0 but I guess, maybe it seems like I need to be looking at where tan is = 0, which is (1,0) and (-1,0)?
I know, this is a tough subject for me.
I don't get a lot of explanations in this class so I have no idea what the bounds or rationale should be.
why do you keep typing tan0
tan(0) is defined, it's value is 0
im getting confused just reading this lol
I know.
I just don't understand how this check is supposed to be helpful for me. It has very little context.
It's in the book, and it's about all I have to rely on.
checking is what you do if you aren't confident with your work and/or if there are certain steps that potentially generate extraneous solution
Yes, I get that, but this particular check. tan0 is defined at -1 and 1. I don't see how that's helpful for me to understand generally why this calculates correctly.
no
for the 5th? time?
tan(0) is defined, it's value is 0
tan(0) is defined, it's value is 0
tan(0) is defined, it's value is 0
tan(0) is defined, it's value is 0
right, what I'm saying is, tan 0/1 = 0. I know that the value is 0.
,calc tan(0)
Result:
0
then why do you keep saying stuff like
tan(0) is defined at -1 and 1
that makes no sense whatsoever
Because it's only defined at -1 and 1.
tan(0) is defined, it's value is 0
What do you think this means?
its undefined at (0,1) and (0,-1)
What does tan(0)=0 have to do with these two points?
i think i get what he says
Means it can produce a legitimate value at that location on a graph. sin/cos = 0/1 = 0. But you can't divide by zero, so at the other locations -1/0 is not a valid value.
No clue. That is why this makes no sense to me.
it's undefined at π/2 by your reasoning, which is right.
Yes, thank you.
That's not a useful way to think about functions
i think you're misunderstanding something here? at no point do you find sec=0 for pi/2 and 3pi/3. you're just plugging x in for sec(x) - 1 = tan(x)
Ok, please help me.
And -π/2
,w plot tan(x)
It isn't useful, perhaps, but I'm not sure how else to because this teacher doesn't actually teach. I'm trying to teach myself. Tutoring has not been helpful either. I promise I'm not trying to just annoy everyone.
ok wait
you don't seem to be using correct terminology which is making it very difficult to understand what you're trying to say
perhaps it would be helpful to start over from the beginning cause everyone's getting confused here
I honestly don't doubt it.
It's better to ask for explanations than to insist on you knowing something you don't
post the original and highlight the first thing line you have an issue with
That's what I've been trying to do, but I keep being told that an answer is an answer. I don't know how else to ask the question.
Ramonov and others here can't both help you and decipher the stuff you're saying at the same time
Told by who
It's this. Generally.
in my mind - what I think I am supposed to do to check is to figure out where sec = 0.
checking is what you do if you aren't confident with your work and/or if there are certain steps that potentially generate extraneous solution
you just plug it in to sec(x) - 1 = tan(x)
if it matches up
it works
if it doesnt
its invalid
the original question that you didn't explicitly post seemed to be solve
sec(x)-1 = tan(x)
squaring both sides of an equation may lead to extraneous solutions so it's a good idea to check here
Ok hear me out. I guess, it's I"m thinking not just of this problem. But other problems in the future I will need to perform the check for. If somehow the value is 5 instead of 0. I'd need to know what sec(5) is, just as an example. Right?
Agree, it is not.
im confused on how exactly you will be applying this check
what is this check for
This particular problem says when I'm applying squares to a trigonometric function, like Ramonov mentioned above.
??
not limited to trig
i think that check is literally just for this, so you check the answer you got from these steps, forget the check after this
(secx-1)^2 = (tanx)^2 for ex.
huh
When you square a function, it says you should check it.
i think the checks just saying
for f(x)^2 = g(x)^2 f(x) = g(x)
at the x value where
ok disregard me
im confused now
but none of this addresses why you kept saying stuff about
tan(0) is defined at -1 and 1
Sure.
Because when I think of where I need to look at values of zero for tan, I'm looking at where on the unit circle that might be.
Unless I am just fundamentally missing something.
tan(0) = sin(0)/cos(0)
i think maybe you're trying to talk about arctan(0)
if you really want to use the unit circle
tan(0) is the ratio of the y and x coord of (1,0)
Yes, but you can't divide by zero. So for places on the circle that cosine is 0, there can't be a tan value, is that not right?
Yes, same page.
no, you need to look at the angle
not the value on the circle
i think you may have a fundamental misunderstanding of how the unit circle works
and that's all you really consider here when checking whether x=0 is a solution here
you don't need to concern yourself with anything else
cos0 is not where cos is 0
you don't need to concern yourself with stuff like pi/2 when that doesn't come even up in the solving process
Ok, question... so, cosx and x=0 is that not where cos = 0?
no
no, i would consider watching a slow and easy YouTube video on unit circle in trig.. it will make it easier
Sure, I was just looking at where sec = 0. I think that's why I thought tan 0 would be in relation to where sec is 0. but it's not, here, obviously.
your seem to have a major confusion with the inputs and outputs of a function
k, thank you.
if you have any other problems just ask again but i suggest you watch a video : )
i think your issue is just a misunderstanding of how trig functions work
cos(x) does not mean you find the angle of where cos = x
functions in general
definitely.
No no, I do know that.
this is sound advice tho
in this check problem though, x was 0, and zero was put back into the problem as sec(0) - where otherwise it would be secx.
I plan to.
cos(t) when t = 0
is cos(0)
which gives the x cood of the point on the unit circle with an angle (anticlockwise from the positive x-axis) of 0
Actually I am glad you said this. I had written down a question about this very thing.
should change a variable there to make it clearer
the unit circle is derived from special triangles (30/60/90) and (45/45/90) so i would suggest looking at that too
and avoid using a variable to represent different things
Alright, yes, agree. Thanks everyone, I really appreciate you. I'll move along!
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I think I just understood what you said. I didn't look at these as functions because they've been calling them equations. not functions. Yes, it makes complete sense within the context of a function. Input a value returns an output value.
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The exercise is: There is a new food which includes three ingredients (A, B, C). The ingredients can have the vitamins B1, B6 and B12 (From the picture you can read which ingredients have how much of the vitamins; The Values are in %, so 0,1 is 0,1%). The new food has a total 0,05% B1 and 0,075% B6
Now we have to find out,
- how big the portion of ingredient B in the new food is
(Output in %) - and what the portion of B12 in the new food is
(Output in %)
My problem is:
There are three equations and four unknown variables. The output for B12 can't be a clear percentage. There will always be a dependency.
Am I wrong ? Can anyone help me find a solution?
Indeed. You could take 1 of C then 1.5 of B, or 0.5 of A then 0.5 of B
They don't give the same amount of B12
Also muss ich hier in Abhängigkeit von einem bestimmten Variablen weiterrechnen.
Könnte ich jetzt sagen: (die erste Reihe - zweite Reihe) 0,1a + 0b + 0,05c - ( 0,1a + 0,05b + 0c) = 0,05-0.075
=> b = (0,025+0,05c)/0,05
Wäre dann "(0,025+0,05c)/0,05" der Anteil B in der neuen Futtersorte?
Just because I answered doesn't mean I speak german fluently though
Oh, I am so sorry 😶 (saw your name and thought you were from Germany)
So I have to continue calculating here depending on a certain variable.
Could I now say: (the first row - second row) 0.1a + 0b + 0.05c - ( 0.1a + 0.05b + 0c) = 0.05-0.075
=> b = (0.025+0.05c)/0.05
Would "(0.025+0.05c)/0.05" then be the proportion B in the new feed?
It's a joke with eigenrocket and eigenzeta
Why do you subtract the rows ? You can't add B1's and B6's like they're the same unit
Just realized my about me can't be clicked
I created a "System of linear equations" and tried to get the variables a, b and c
A, B and C are more like vitamin vectors though
Operations on columns make sense
Rows, not so much
This is what I thought.
Now you could use Gaussian elimination and try to get the value of the variables
Why isn't it allowed to say I - II or II - I and then solve for b
Yeah actually
"yeah actually it is"
Yeah you can
hmm, ok!
Thx for your help
My physics brain taking over
When it shouldn't
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I don't understand what you would do here
arg(z-8) - arg(z-2) = pi/2
The points 8,0 and 2,0 are on x axis so how can there be an angle between them
also the angle pi/2 would just be a verticle line upwards so how would that work
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(A) ∀x∃y (x ≥ y) (B) ∃y∀x (x ≥ y), x and y are natural numbers
@vagrant pasture Has your question been resolved?
@vagrant pasture Has your question been resolved?
A. For all x there is some y such that x >= y B. For some y all x are >= y
is that what it's saying
shouldn't b entail a too
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I have kinda no idea how to start this problem. I recognized it as a M/G/1 queue with arrival rate 8 and service rate 2k but I'm not sure what to do with that information
I'm thinking for part (a) by PASTA principle we can find $\pi(0)$ to find the time that the professor is idle and the stationary distribution of the number of letters to be written is just $1 - \pi(0)$
heheitsop
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I have a questoin on the following probability exercise
you have to pick one of these boxes with closed eyes and then pick a ball out of that box you've chosen previously. Whats the probability to pick a white ball
now the point I am struggling with is, I need the probability of each box, and my brain tells me its 1/3 each but my math tells me that bs because then I always end up with false probabilites
what "false probabilities" are you ending up with, and how?
so lets say we name the boxes B1 B2 and B3
then I calculate
P(W | B1) so the probability to pick a white ball given I picked box 1 and that probability is greater than 1
which, afaik, is bs
P(W | B1) so the probability to pick a white ball given I picked box 1 and that probability is greater than 1
how is that happening?
yeah P(W&B1) is not 5/6
P(W|B1) is 5/6
P(W|B1) is the probability to pick a white ball given that you already know your box is #1
it's the probability that you get the first box AND draw a white from it
hm I always thought of it like, what are the elements that are both in W and B1
P(W n B1) would actually be 5/18 here
there are in fact 18 balls to choose from that are all equiprobable
(because all the boxes have an equal number)
I think I am having something mixed up in my head
one second
this is another exercise I did previously
we have 640 females and 360 males
and 15% of F smoke and 20 % of M
now that would make 96 female smokers in total
R = smokers?
here I imagined it like we have a pool of 640 females and I intersect it with those who smoke
yes in german it is raucher
and now here it was totally clear to me what the intersection would be
but with the boxes and balls I am a bit confused
or no, I think I can work with that. I'll just accept this for now. 5/18 is not too incomprehensible for me.
thank you for your help !!
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\
i've seen this exact question posted before I think
but the topic is combinatorics/counting
Do you know any youtube channels that explain good pls?
Probability - https://www.khanacademy.org/math/statistics-probability/probability-library
Counting and combinations - https://www.khanacademy.org/math/statistics-probability/counting-permutations-and-combinations
I don't know of any personally, sorry. I don't know if Khan Academy has any videos for it. You could try seeing if Harvard's course STAT110 has any videos/lectures online
well, there you have it lol
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Hi can someone help me with question 1
do you know about the unit circle?
if u say u know about the unit circle, what value of angle (x) has their cosine value = cos ( pi/3 )?
find those x that lies within the interval u want
60 degrees
what else?
1/2
there are more than one such x if we consider from 0 to 2pi
do u know how sine and cosine function value oscillate between -1 and 1?
Oscillate?
No
Yes I know these values I got confused with this
sine is essentially the y value and cosine is the x value in the graph, (tangent = y value / x value)
Ye ye I know that
so in what degree/ radian that share the same cosine value as pi/3
other than pi/3
just between 0 to 2pi
5pi/3
yes
and if we consider larger interval or in real number every 2pi the pattern repeats
ur question ask for 3pi/2 to 2pi so udh to consider others but in generally there is more roots
Oh okay so I’m essentially using the value given and I’m just adding/subtracting 2pi until it fits into the domain
/restriction
not rly adding or subtracting 2pi, it depends on the function
sine, cosine, tangent each has different pattern
for example for tangent, u just keep adding/subtracting pi
Wouldn’t it be easier just to subtract the maximal value from pi/3
$cos (x) = cos (k)$,
$x = 2n\pi \pm k$
wannabe
And even in the next question for sine pi/4, I would just do pi-pi/4
not like this
Wdym
if next question change to cos x = cos (pi/4), find x in (pi/2, pi)
there would be no solution
i mean u can argue that ur hw wouldnt have "no solution" as ans
Where have you gotten that from
and what if ur question doesnt give "nice" interval, like sin x = sin(pi/4), find x in (2pi, 4pi)
just some random question
Oh okay
Ye fair enough
Because even if you try to fit that into the given domain
You cannot
Based on adding/subtracting pi
its better to understand the concept or pattern behind instead of, trial and error for each npi +- k
like if the range get rly large, u dont rly want to find them one by one
So in terms of understanding this formula how does it work
the easiest way it is to visualize it like how sine, cosine, tangent value changes if u dw to memorize the formula
So what’s n?
any integer
For example?
cos x = cos (3pi/2) = 0
x = 2npi +- 3pi/2 for all integer n
..., -3pi/2, -pi/2, pi/2, 3pi/2, ... are all possible ans for x
then just pick those that lie in the interval u want
if $sin (x) = sin(k), k\in[-\pi,\pi]$, then
$x = n\pi + (-1)^{n}(k)$
wannabe
Yep yep
So I’d just let n = -3 to 4 (for example)
Depending on my domain
then I would find the values of x that locate within my domain
if $cos (x) = cos(k), k\in[0, 2\pi]$, then
$x = 2n\pi \pm k$
wannabe
not sure how u get n=-3 and n=4
yea sure
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can someone confirm this didnt get integrated properly
as the two 1/2's didnt get integrated properly and ones missing
I guess its correct
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,rotate
@warm oriole Has your question been resolved?
Which one is the problem you need help with?
He said 8
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I’m confused on whether to use shell or washer method here ?
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Did I do this well?
Q: Adam lives in Bocianowo, and his friend Bartek - in Zabno. Adam made an appointment with Bartek in Zabno at 18:00. He left Bocianowo on a scooter at 17:20. Adam's average speed was 25 km
On a square grid, Adam shows a diagram of the route he is following.
What time did Adam arrive to meet Bartek? Save calculations.
Distance from stawisko to bajorko can be calculated using pythagoreas theorem
like this
@zealous mason Has your question been resolved?
no
it is 4 km up and 3 km right
so a 3 4 5 right triangle
distance between stawisko and bajorko is 5
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if ur integrating ln x all squared
Seems like a valid approach
That works as well, cause you will then have to integrate 2lnx
In fact that is the better way
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Two numbers, x and y are positive integrals. Which numbers are x and y if, x/3 + y/11 = 31/33
I've first gotten a common denominator: 11x/33+3y/11 = 31/33
(11x+3y)/33 = 31/33
So 11x+3y = 31
yea
How do I solve
Theres only a few cases
just try it one by one
is that the only solution
no
can't I do it another way
but that's the easiest
yea ofc
You know that 0<=x<=2, because if x is 3 then it's bigger than 33
You could use a greedy algorithm here
Find out how many 11s go into 31
Then find out how many 3s go into whats left
Yes
just a question, when do I know when I should use a greedy algorithm
In general its a good starting point for question like these
You might have to modify it a bit
got it thanks mate
Like for instance if you instead had 11x + 3y = 23
Then two 11s would give you 22, and 23-22=1, and 3 doesnt go into 1.
So then try one 11, then 23-11 = 12, and 3 goes into 12
So that works
It's an iterative process
but it works for all equations with two variables?
Yeah you'll eventually get the right answer, it's an algorithm
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how come
-29 mod 3 = -1 and not -2
I understand there is some formula or whatnot but I can't understand it logically
if 29 mod 3 = 2
because you can fit 3, 9 times inside
then have remainder of 2
are you sure you're not inadvertently calculating -(29 mod 3) as opposed to (-29) mod 3?
I can't really understand -29 mod 3
conceptually
also, beware that different programming languages (and maybe different calculators) will have different conventions for this
I can understand with the formula and the explanation
a mathematician would expect the result to be 0, 1 or 2
but computer people have their own weird desires
ah gotcha
I'm learning modular arithmetic for math competitions
not interested in programming or any of the sort
what source is telling you that the answer is -1?
Khan Academy
interesting
wait nevermind
it says
1
but I still don't understand it conceptually
how it's 1
ah that's correct then
the idea is this:
add or subtract multiples of 3 to -29 until you get a number between 0 and 2
in this case:
-29 + 3x10 = 1
yep exactly
gotcha
and an alternative way to do it
would be
-29 + (9*3) = -2
-2 mod 3
subtract a 3
-2 = -3 + R
R = 1
yep that works too
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sure, cheers
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Stephen
riemann
huh
I am as lost as possible right now with the problem
<@&286206848099549185> (sorry for the second ping but i really really need help)
@wintry radish Has your question been resolved?
😭
Show your work, and if possible, explain where you are stuck.
I have no idea where to start
its asking for the rate of change at some time
how do you find the rate of change at a time
you derive?
the word derive isnt a calculus thing
derive just means like figure out
i think you mean differentiate
oh yeah
and how do we find the rate of change at a specific time t
it would just be dD/dt and you plug in some number for t
dD/dt is some function of t
uh huh
so do you know how to take these derivatives
sure
what is dD/dp and dp/dt
,w derivative of 7225/(3p+13)
thats what i got
okay cool just checking
cool
so t is 36 and p is 24?
so then that means that the rate of change is like -9/4?
that makes no sense how can there be negative blenders
wait slow down
ok
I plugged the values
which is just the instantaneous slope of demand
wha
so that can be negative
any decreasing function that has a positive value has negative slope
i suppose it makes sense now then
okay cool
I have another question i was wondering if you could help with this one too
or do i need another help channel
helloooo
you can just ask it here
!volunteers
Helpers are just people volunteering their time to help you. Be polite.
watch some videos on related rates
these are honestly annoying to explain every time
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I have 2 more questions at the end of this practice test that I could not solve.
The first one, I was able to understand how to get the final answer
just now how to elaborate
let me get the table
I understand that X can represent the "questions correct"
but I wanted to know a better way to word this
One way to consider it is you have a binary choices. A coin flip has either H and T. True or false has either T or F. Yes no has yes and no. 2 options only
When you flip the coin n times and you want x of an outcome, you have a probability of nCx/2^n. The 2^n is there because we have 2 options with n trials.
open a new channel #❓how-to-get-help
Sorry I'm new
np! Head to that channel and it'll explain how to get help 🙂
You have the same number of choices for each style.
For example, if you were to ask 10 people which ice cream they'd prefer: vanilla or chocolate, you are only providing 2 choices.
I do understand all that, still trying to figure out how to write it out in a way which gets the full 3 marks
Or 2 marks, I assume the final answer is the last mark
I would use the term "binary choice". You get one or the other. Something like: "When flipping a coin, we have a binary option between heads and tails. Since answer can only be true and false, this is also only two options, so you would use the same combination of 10Cn/2^n for each P(X=x)". To find out the probability of getting five or more questions, we get: (10C5 + 10C6 + 10C7 + 10C8 + 10C9 + 10C10)/2^10 = (252 + 210 + 120 + 45 + 10 + 1)/1024 = 638/1024 ≈ 62.34%"
or something to that degree
Thank you, this really helps a lot.
yep np!
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wwwwwwww
What are the answers
!nosols
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
then how would I find the answers
brute force
What
find all the zeroes
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. help
How can the common denominator be xy when the numerator only has an x???
not sure what you mean
they employed the idea of multiplication by 1 to simplify the fraction
in this case they multiplied the numerator and denominator of the big fraction by the lcd which in this case is xy
Azka Aska
no
Ok I got it now
That was easy
Could you by any chance help me with one more thing that's been stumping me for 3 hours now?
I have no clue what to do next
okay
okay i did it out
good
from here
simplify numerator on your left hand side
2(x-2)-x =??
yes
and now you can cancel out x-2 on both sides
leaving you with x-4/x=-4/(x+2)
Alright, got that
now cross multiply
Can I do that even though they're on opposite sides of the = sign?
yes that's precisely the case to cross multiply
Would I be left with a denominator, or is (x-4)(x+2)-4x all that I'm left with now?
Would my final answer be (x-2)(X+4) = 0
Which gives me teh solutions x=2,-4?
It is
You actually made my life a million times easier
thanks so much
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How did we end up with (a.c)(a.c) here?
Shouldn't it be [(a.c)a].c
they're the same
😮
an easy way to see that is to let some other variable t=a.c
then [(a.c)a].c = [ta].c = t[a].c = t(a.c)
Assuming that t is some scalar right.
if a.c is dot product of a with c, then a.c is a scalar yes
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what is the highlighted part in set symbols ??
p=|(X u Y u Z)'| I guess
yeah
the cardinality (number of elements) would then be denoted with | |
so p = |(X union Y union Z)'|
oh ok
thx
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A coin is tossed 3 times then the total possible number of events is ( 2^3 = 8 ).... right??? (just confirming)
@lean otter Has your question been resolved?
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there is a x,y function contained in this area (im working with joint functions, statistics) and im trying to get the y function from that function, how do I do this?
lets call that function idk f(x,y)=1/2, just to make it simple?
I know that I need to take integrate x away from that joint function but idk how
I tried integral(1/2) dx from x=y+1 to y-1 but that doesnt seem right at all
I would let x go from -1 to 1 and y go from x-1 to x+1
But what you said seems to be right as well, except it should be from y-1 to y+1
how come? I thought area under y-1 to the x axis is positive minus the area above y+1 to the x is negative, so you get the total area?
Im more confused with the actual solution:
im struggling to understand what he means by taking the integral from x=y-1 to 1, what does this exactly mean?
also the function was a bit more complicated than 1/2, I dont really care about solving the integral, im stuck understanding the boundaries and what the integrals are trying to do
First they integrated the area above x-axis and then the one below x-axis it seems
But that is a really weird step
What's the final answer btw?
Okay nvm it's not
Here is what letting y go from -1 to 1 and x go from y-1 to y+1 yields
als= if , elders = elsewhere
So we are just doing the inner integral? (the one with respect to x)
I dont really understand what you mean by inner integral tbh but we integrate that joint function with respect to x, over the domain of x where y lies
Assuming the region given is that
yeah it doesnt seem right, especially compared to the solution
Oh I think I see what they did now
Like I was thinking, they simply cut the shape in half
And integrated each separately
wait I dont get how for example for y=0to2, that integrating from x=y-1 to 1 gets you that purple area
wont you get that 1 square minus the area under y-1, that you form the triangle above that purple triangle/area?
I was looking online for similar problems and it seems I dont understand this one thing
solution was
and I again thought that it was 0 to y
so why are we doing the 'opposite'?
Why would you consider completing the square there though, it doesn't help when integrating a multivariable function
It's from y to 1 because the triangle is located between x = y and x = 1
Not between x = y and x = 0
Then it would be the other triangle yeah
wait Ill just show what I mean
, nvm I see my mistake lmao, Ill just quickly redo the 'opposite'/correct way and see how that works and ill close the thread, thx btw
altho I think I still have a question, why split the 2 areas up from y -2 to 0 and y=0 to 2?
wait I take it back
but isnt this located between x=0 and x=y?? so if you do y to 1, dont you get this one where you do green minus purple?
Why are you subtracting purple from the green?
isnt that what we basically do when integrating from y=x to, oh, I might see my mistake, wait a sec
ok yeah now Im confused how that blue/purple area gets defined from x=y to x=1?
please explain like im 5 lmao
Can you name the lines between which that triangle is located?
y=0, x=y and x=1
Right, so it's between x = y and x = 1 (and y = 0)
yeah
Meaning x goes from y to 1
yes
so what do you get if you do 1 to y
and so we split this in 2 because there are to many lines for an integral with 2 boundaries?
Yeah
You can do it as one integral but
That would involve min and max functions
Which require dividing the shape anyway
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i am given a function $k(x_1, x_2, x_3, T)$, each argument except for T has an absolute error value. T is given as 5 possible values and i can calculate k 5 times. I can also calculate the gaussian error (based on error propagation of uncertainty using $ \delta k = \sqrt{\sum_i (\frac{\partial k}{x_i} x_i)^2}$).
I need to present k as its average over thsoe 5 possible T values, how can i get the guassian error of the average then?
derdotte
I remember that i can not simply average all 5 errors as that isnt how this works. Do i take the largest error of the prior k and add/subtract it to δk?
@eager prawn Has your question been resolved?
<@&286206848099549185> anyone mind helping?
Yes
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$$\forall n\in Z \text{ : } n\ge Z\implies n^2\ge 4$$
BachToReality
I was watching a lecture online, where the professor gave this as an example for implication.
How does this hold true, for say n=0 or n=1?
1 >= 1 implying that 1*1 >= 4? Am I doing this right?
This is wrong in a lot of ways
Did you mean n >= 2?
are you sure you didnt misread what they wrote?
Are you sure you wrote it right
n>=Z makes no sense at all
n>=2 would make much more sense and additionally would make the implication true
You might be right.
I would've mistook it for 2. It's a lecture from MIT OCW, guess my wet dream of disproving MIT profs will never become reality.
Thanks for your help!
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Confused on what this is asking for
Do you know what an augmented matrix is?
No
Then that is where I suggest you start looking into
It wants that system of equations in augmented matrix form
This is useful
https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-representing-systems-with-matrices/a/representing-systems-with-matrices

