How about and? not( p and q) is false only if both p and q are true.
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How about and? not( p and q) is false only if both p and q are true.
kidtamagotchi
Member
Thanks! I'll try it out!
Partial Gamification
Banned
t: I am tired
h: I am hungry
c: I can concentrate
~ : not
=> : implies
V : or
A : and
condition: t V h => ~c
given: c
thus : ~(t V h)
c => ~(t V h)
~(t V h) <=> ~t A ~h
c => ~t A ~h
The notation is not standardized, ¬ is a better choice for not than tilde. You could strike the assertion:
Just got my grade back for Calculus and got a Solid B. Thank you to everyone that helped me out this semester.Now onto Calc 2 during summer.
Congrats and good luck tot!
Lion Heart
Banned
Gr 12 shit here, lets go.
I have to figure out the zeros, and we learned the synthetic division method. When its not easily solvable with real integers, I don't know how to proceed. The answers are (rounded) x= -0.7 , 0.88, 0.9.
I have to figure out the zeros, and we learned the synthetic division method. When its not easily solvable with real integers, I don't know how to proceed. The answers are (rounded) x= -0.7 , 0.88, 0.9.
Hey GAF I got a Calc final tomorrow and I need some help
So, would this be:
a) 0 because limit from 1 to 1 = 0 area
b) (1+x^2)^1/2 (just subtitute x because of the fundamental theorem)
c) Derivative of the above function? So (1/2(1+x^2)^-1/2)(2x)
Yep... pretty much... On the first look I was gonna trig sub but then I noticed those other things...
Gr 12 shit here, lets go.
I have to figure out the zeros, and we learned the synthetic division method. When its not easily solvable with real integers, I don't know how to proceed. The answers are (rounded) x= -0.7 , 0.88, 0.9.
Th is not something you're gonna get on a test... .7 x .88 x. 9 isn't equal 6... that's gotta be an imaginary somewhere...
Edit tested it out on Wolfram Alpha... it's horrible complex numbers. Don't worry about it...
Lion Heart
Banned
DEATH;57802464 said:
Yeah, you're right but I had questions similar to this multiple times in the chapter review. Is there not a way to figure it out without a graphing calc?
You can put in numbers and note when it is positive and negative to narrow down on the roots. For example f(0)<0 and f(1)>0 so you know a root is between 0 and 1. Then try 0.5 and you can narrow it down again
DEATH;57802464 said:Yep... pretty much... On the first look I was gonna trig sub but then I noticed those other things...
Th is not something you're gonna get on a test... .7 x .88 x. 9 isn't equal 6... that's gotta be an imaginary somewhere...
Edit tested it out on Wolfram Alpha... it's horrible complex numbers. Don't worry about it...
Awesome, thanks.
No, plugging it into V just returns V=32. Plus, you don't want to minimize or maximize the volume, you know the volume must be 32. You want to minimize the amount of material used, so you can use the relationship for surface area and substitute into that to make it in terms of one variable. Then you take the derivative and equate that to 0.
Partial Gamification
Banned
Gr 12 shit here, lets go.
I have to figure out the zeros, and we learned the synthetic division method. When its not easily solvable with real integers, I don't know how to proceed. The answers are (rounded) x= -0.7 , 0.88, 0.9.
You could find the critical point(s) of the graph and plug the values in to find where the graph's slope is zero. From there, test points on either side and get a feel for the graph's shape and location on the plane. just really adding to f0rk's suggestion.
I saw the roots and tried to see if the classic Cardano's Method would hold up and, well, this is was third attempt (I multiplied by negative-one) and I'm going to leave it:
Regarding the roots, the graph shows there are two real roots, but they are fugly. The function has been altered (multiplied by negative one) but the roots are the same, each root is written three ways.
This is a tough one, what other methods have you used in class?
The complex roots have to come in pair, since all coefficients are real. There can't be single complex root.
With google calculator, it clearly shows three roots, they are actually -0.7 , 8.8, 0.9 - a typo in the OP.
google calculator:
https://www.google.com/search?clien...r.r_cp.r_qf.&cad=b&sei=oGCUUcDCNarR0gHxiYCoDw
Partial Gamification
Banned
All three roots had imaginary units, second and third cancel being simplified in Maple.
Try it in a CAS program then simplify and approximate the solution.
edit: the third's imaginary component should cancel but I don't know how to get it out of the given expression.
edit2: I think the point was missed on my part. A cubic could have one complex root, imagine the graph crossing the x-axis, having a critical point above or below the horizontal axis, and then having another critical point on the x-axis. Two real roots, but not in the given question -this is possible, yes?
Not according to complex conjugate root theorem. When it has a critical point on the x-axis, that means it has two identical real roots.
Partial Gamification
Banned
Yeah, not that specifically, but it hit me that I was talking of an instance of repeated roots (the critical point on the x-axis). I had this bugging me all day long. Part of me laughs at the error and part of me is just broken. All and all, thanks for highlighting the Theorem.
triplestation
Member
i just came back from my algebra final
i got a perfect score!!!!!
i got a perfect score!!!!!
Nice, congrats!
Buggy Loop
Gold Member
Riemann integral help!
This thing is killing me. I have no fucking clue. Just for reference i had 94% in my final of derivative, but this is just a beast of its own. Doesnt help that the teacher sucks, his book he made us buy for 50$ sucks and that its a summer course.
Integral -2 -> 1 x^2 dx
Im actually finding the derivative method to be MUCH easier than Riemann's, i would know that its (X^3) /3
and that (1^3)/3 - ((-2)^3)/3) = 9/3 = 3
But for the life of me, i cant with riemann. Here's my procedure and perhaps someone can point where i screw up
Delta X = (b-a)/n = (1-(-2))/n = 3/n
Xk = a + (k(b-a)/n) = -2 + (3k/n)
Lim of n-> infinite 3/n sigma (-2+(3k/n))^2
Lim of n-> infinite 3/n sigma 9k^2/n^2 - (3/n)*4 sigma(1)
Lim of n-> infinite 27/n^3 sigma(k^2) - 12
And from there, it does not go down well toward a 3 for answer. I end up with 9 - 12. There must be something weird with my -2 from the xk equation. Typically the other exercise or examples i've seen were with a=0.
Thinking about it, Lim of n-> infinite f(xk)*deltaX
f(xk), would it actually end up being (3k/n)^2 -2 ? Then i would get 9-6=3.
But that doesnt make sense to me, f(xk) with x^2. It ends up putting everything that xk is squared, including the -2
...
Holy shit nevermind! I didnt develop (3k/n - 2)^2 rightfully. Such a newb mistake. I realized that while taking a shower, i guess i needed to wake up. lol
This thing is killing me. I have no fucking clue. Just for reference i had 94% in my final of derivative, but this is just a beast of its own. Doesnt help that the teacher sucks, his book he made us buy for 50$ sucks and that its a summer course.
Integral -2 -> 1 x^2 dx
Im actually finding the derivative method to be MUCH easier than Riemann's, i would know that its (X^3) /3
and that (1^3)/3 - ((-2)^3)/3) = 9/3 = 3
But for the life of me, i cant with riemann. Here's my procedure and perhaps someone can point where i screw up
Delta X = (b-a)/n = (1-(-2))/n = 3/n
Xk = a + (k(b-a)/n) = -2 + (3k/n)
Lim of n-> infinite 3/n sigma (-2+(3k/n))^2
Lim of n-> infinite 3/n sigma 9k^2/n^2 - (3/n)*4 sigma(1)
Lim of n-> infinite 27/n^3 sigma(k^2) - 12
And from there, it does not go down well toward a 3 for answer. I end up with 9 - 12. There must be something weird with my -2 from the xk equation. Typically the other exercise or examples i've seen were with a=0.
Thinking about it, Lim of n-> infinite f(xk)*deltaX
f(xk), would it actually end up being (3k/n)^2 -2 ? Then i would get 9-6=3.
But that doesnt make sense to me, f(xk) with x^2. It ends up putting everything that xk is squared, including the -2
...
Holy shit nevermind! I didnt develop (3k/n - 2)^2 rightfully. Such a newb mistake. I realized that while taking a shower, i guess i needed to wake up. lol
I'm doing this trig equation and I am getting the correct answer, but my book is telling me there is another correct answer which I don't know how was reached.
1+Sec x = tan x
It wants all solutions on [0, 2pi)
Here's what I do:
1+sec x= tan x gets squared on both sides giving me
(1+sec x)^2 = (tan x)^2 which becomes
1+ 2sec x + sec^2 x = tan ^2 x and I use a trig identity on the right side of equation to get
1+ 2sec x + sec^2 x = sec^2 x -1 so there is "sec^2 x" on both sides, which cancel each
other out, leaving
1 + 2sec x = -1
2 sec x = -2
sec x = -1
arcsec -1 gives me a final answer of pi. Great, but the problem is my book is saying 0 is also a correct answer.
When I check 0, it works, but I don't see how I could have come up with that while solving the equation.
1+Sec x = tan x
It wants all solutions on [0, 2pi)
Here's what I do:
1+sec x= tan x gets squared on both sides giving me
(1+sec x)^2 = (tan x)^2 which becomes
1+ 2sec x + sec^2 x = tan ^2 x and I use a trig identity on the right side of equation to get
1+ 2sec x + sec^2 x = sec^2 x -1 so there is "sec^2 x" on both sides, which cancel each
other out, leaving
1 + 2sec x = -1
2 sec x = -2
sec x = -1
arcsec -1 gives me a final answer of pi. Great, but the problem is my book is saying 0 is also a correct answer.
When I check 0, it works, but I don't see how I could have come up with that while solving the equation.
Lion Heart
Banned
Solve algebraically:
1/x^2-2x-7=1
God this should be so easy.
nvm, I got it, thanks anyway.
1/x^2-2x-7=1
God this should be so easy.
nvm, I got it, thanks anyway.
It's Mindset bro... MINDSET...
Note that at x = 1, f'(3x^4) equal f'(3), and f'(3) is given, so you don't need the generic f'(x).
Yep, you're almost there:
We have y=f(g(x))=f(3x^4)
therefore y'=f'(g(x))*g'(x) - via the chain rule
we also know f'(3)=-2 (that is 3x^4=3 => x=1)
thus y' at x=1 is y'=(-2)*g'(x=1)
I have this homework problem regarding Linear Algebra, it reads...
Show that the Matrix B commutes with the Matrix A. Show explicitly that B is diagonal in the same basis in which A is diagonal.
B=
0 1 1
1 0 1
1 1 0
A=
2 0 0
0 1 1
0 1 1
I know they commute because AB = BA via matrix multiplication, but what does it mean that they are diagonal in the same basis? They don't have the same eigenvalues? Unless I'm doing something worng
Show that the Matrix B commutes with the Matrix A. Show explicitly that B is diagonal in the same basis in which A is diagonal.
B=
0 1 1
1 0 1
1 1 0
A=
2 0 0
0 1 1
0 1 1
I know they commute because AB = BA via matrix multiplication, but what does it mean that they are diagonal in the same basis? They don't have the same eigenvalues? Unless I'm doing something worng
I have this homework problem regarding Linear Algebra, it reads...
Show that the Matrix B commutes with the Matrix A. Show explicitly that B is diagonal in the same basis in which A is diagonal.
B=
0 1 1
1 0 1
1 1 0
A=
2 0 0
0 1 1
0 1 1
I know they commute because AB = BA via matrix multiplication, but what does it mean that they are diagonal in the same basis? They don't have the same eigenvalues? Unless I'm doing something worng
(It has been a while since I did linear algebra.. but I'll try to answer you)
Your matricies are both 3x3. We know that diagonal 3x3 matricies can be represented by a basis consisting of 3 matricies, so dimension 3.
Also, we know that if A is a square matrix of order n and If A has n distinct eigenvalues, then A is diagonalizable.
So just check how many Eigenvalues A and B have. If they both have 3(the dimension of your basis) then they are both diagonal in the same basis.
The Eigenvalues of A are 0 and repeating 2
The Eigenvalues of B are -2 and repeating 1
Am I supposed to get 3 different Eigenvectors?
Actually the second matrix only has 2 eigenvalues, but it's still diagonalizable because it has a full set of linearly independent eigenvectors. If P is the matrix whose columns are the eigenvectors of B, then B is diagonalizable with respect to the basis given by the columns of P. Or in other words, P * B * P^-1 is a diagonal matrix (whose entries are the eigenvalues of B). You just have to find P and show that P * A * P^-1 is also a diagonal matrix.
Although, I tried doing it with an online calculator and I'm getting something which isn't diagonal for P * A * P^-1, so the question might be wrong, or maybe I'm doing something wrong.
It shouldn't matter if your eigenvalues repeat. It would just lead to a diagonal matricies that looks like:
-2 0 0
0 1 0
0 0 1
And
0 0 0
0 2 0
0 0 2
Still both diagonal in the same basis of dimension 3.
Yeah this looks good, I shouldn't do math while half asleep.
I have a terrible teacher, I watched a youtube video of how to get P, but he goes too fast and he erases as he goes... how do you get 3 eigenvectors from 2 eigenvalues for B?
You can have more than one linearly independent eigenvector for a given eigenvalue. For instance, if the matrix (A - lambda*Identity) is
1 1 0
1 1 0
0 0 0
then both <0 0 1>^T and <1 -1 0>^T are eigenvectors
If you were using the Eigenvalues that I posted then it's wrong, found my error... one stupid (-) sign. The Eigenvalues for B are 2 and repeating -1. Thanx
I have another homework problem:
Show for an operator A that the matrix representation of the Adjoint, A-dagger, is given as the complex conjugate of the transpose of the matrix representation of the operator A, ie A-dagger = (A-transpose)*
A=
a b
c d
A-dagger=
a* c*
b* d*
A-Transpose=
a c
b d
Just show that the Det are = ?
Show for an operator A that the matrix representation of the Adjoint, A-dagger, is given as the complex conjugate of the transpose of the matrix representation of the operator A, ie A-dagger = (A-transpose)*
A=
a b
c d
A-dagger=
a* c*
b* d*
A-Transpose=
a c
b d
Just show that the Det are = ?
Haven't done this in a long time so I'm stuck on what I suspect is an easy problem.
I have two random variables, X and Y, in which, X ~ exp(1/2) and Y|X = x ~ exp(1/3x).
I first have to calculate E[Y], which is 6 (So E[X] = 2 and E[Y] = 6). I then have to calculate Var(Y), which is where I get stuck. I'm apparently supposed to move things around and get to Var(Y) = 9(2Var(X) + E[X]^2), which I have no problem with. Knowing the answer, I know Var(X) = 4, but I have no idea how to get there, as I'm used to having a sample of data to get the variance.
I could also get the answer by calculating what E[X^2] is, but I don't know if that's possible with the info I have.
I have two random variables, X and Y, in which, X ~ exp(1/2) and Y|X = x ~ exp(1/3x).
I first have to calculate E[Y], which is 6 (So E[X] = 2 and E[Y] = 6). I then have to calculate Var(Y), which is where I get stuck. I'm apparently supposed to move things around and get to Var(Y) = 9(2Var(X) + E[X]^2), which I have no problem with. Knowing the answer, I know Var(X) = 4, but I have no idea how to get there, as I'm used to having a sample of data to get the variance.
I could also get the answer by calculating what E[X^2] is, but I don't know if that's possible with the info I have.
for a stochastic variable X that is exp(1/n), E[X] = n, V[X] = n^2.
i am in dire need of answer to a math problem. i have a 100% streak of completing my online homework this quarter but i'm stuck on this problem and my streak is about to be broken. i feel like i know how to do it but the system is telling me it's wrong.
rationalize the denominator
10/9*sqrt(2)
so ten divided by nine then two under a radical sign, i'm not sure how to express that exactly in text.
i'm getting 10*sqrt(2)/36. (multiplying both sides by square root of 2)
the thing says i'm wrong.
IM DESPERATE. what gives!?
rationalize the denominator
10/9*sqrt(2)
so ten divided by nine then two under a radical sign, i'm not sure how to express that exactly in text.
i'm getting 10*sqrt(2)/36. (multiplying both sides by square root of 2)
the thing says i'm wrong.
IM DESPERATE. what gives!?
For the denominator sqrt2 * sqrt2 is 2. 9*2 is 18, So the 36 should be 18.
cpp_is_king
Member
Also it might expect the answer in lowest terms.
Lion Heart
Banned
I missed the lesson for this topic so I don't know how to solve the following (easy one):
cosx + 0.75 = 0
the answers are x = 2.42, x = 3.86
cosx + 0.75 = 0
the answers are x = 2.42, x = 3.86
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