You are not logged in.
Let T : V --> W be a linear transformation with ker(T) = {0}
Suppose u1, u2 ∈ V are linearly independent.
Prove that T(u1) and T(u2) are linearly independent vectors in W.
Just not sure how to construct this. Any help would be great, thanks.
Use codecogs, I sent you the link already.
cool, thanks. Are you able to help me with this question?
dy + y³x+y=0
dx
solve
Looks innocent but I can't figure.
Sorry for all the DE questions, I have a russian lecturer who I can't understand so I need to learn somewhere
Hi;
Is this what you want?
or
with y(1)=1?
the 2nd one sorry. I don't know how you do those nice symbols haha.
Integrate both sides which will eliminate the differentials dx and dy.
Thanks a lot. Would you be able to help me with my initial value problem that i just posted?
dy + 3y + 2 = 3x , y(1)=1
dx x
Not sure how to go about this one.
Integrate dx or 1 *dx
xcos(v) xcos(v)
any idea?
The DE is separable:
Now can you integrate both sides?
Ok so what does it mean by solving explicitly for the y differential equation? do i only intergrate the side with the y or what?
Solve explicitly for the y differential equation.
dy = 3x^2(1+y^2)
dx
I don't really know what this means. Any help means a lot.
n+1
(-1 )^(n-1) * 2n+1
what happens as n goes to infinity?
My thoughts are that it oscillates between near 0.5 and -0.5 but never hits either. does this mean it diverges?
Let Q be the matrix: 1 2 k
0 -1 2
k 2 1
Find all real values of k such that the columns of Q do not form a basis of R3.
My thoughts on this: k appears in the x and z row. It seems to me that the 2 known elements in the x and z row already span the entire xz plane. Therefore I think that any value of k other than zero is the answer but i'm really not confident.
Would appreciate help, thanks.
Matrix P =
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
Can I just assume that the det=1 because it's just a re-arrangment of the identity matrix? Or does the mix up of order affect the determinent? thanks.
Test for validity in LPC
∃x(∃yPy ⊃ Px)
I can do more complex ones, but it’s the simple ones like these which I’m not sure about.
Show that it is valid to infer that, provided there are some stones, if every god lifts every stone, then there is a stone which every god lifts.
So this is what I get as the schema
Px = x is a God
Qx = x is a stone
Rxy = x lifts y
∀x(Px ⊃ ∀y(Qy ⊃ Rxy) ⊃ ∃y(Qy ∧ ∀x(Px ⊃ Rxy))
1 0 0
My problem is that there is no weak quantifier. Usually I solve these by finding the weak quantifiers and creating a-rows and b-rows....however I have no idea what to do when every quantifier is strong.
Help???
You will get that if you integrate both sides
maybe you can show me?
dT/dt = -k(T-T1)
The question then states that this means that T(t)=T1+Ce^(-kt)
I don't see how they got that...can you help please?
What do you have to solve for?
who would know? i've given you all the information that i've been given it just says find the general solution.
hi mrpace
Let's just check that question. Is it:
where X is a 2 by 1 matrix ?
So two variables, t and x ?
Bob
yes that's right. please tell me you can help?
X=
| 1 1 | | cos(t) |
|-1 1 |X + | e^x |
any help is appreciated greatly. Thank you
Sam and Maria are playing a game where a dice is rolled. If it lands a 1 or a 2, Sam wins $10 off Maria. If the dice lands a 3,4,5 or 6, Maria wins $10 off Sam.
Seems like a great game for Maria right?
But what if Sam had an infinite amount of money, and Maria has a finite amount. ($100 for example.) What if these players played this game an infinite amount of times. (let's assume they are immortal and they never lose the dice.)
Does Maria eventually go broke?
Consider the following equation:
x¨ + x˙ + x = 1.
Determine the equilibrium solution and investigate its stability.
not too sure on this stuff. any help is appreciated.
Every vertex of a certain tree has degree 1 or 3. If n vertices have degree 1, how
many have degree 3?
how can i show this?
Consider the following situation, where P is a two-place predicate:
Universe = a, b
P-list
Paa
Pbb
Calculate the truth value of Ex(VyPxy v VyPyx) in this situation.
please note there should be a line under P-list
also note, E=there exists
V=for all
v=or
any help is welcomed, thanks.
(a) Vx(Px U Ey(Qy ^ Py)) = Px
(b) Ey(Ex((Px v Qx) ^ Px)) U Py)
note: let V= for all, U= that symbol that looks like a U rotated anti clockwise quarter of a turn, E=there exists, ^=and, v=or
and the equals sign in the first line is meant to be 'logically equivalent to.
to me, all the variables look bound, but something tells me that's not right.
given that 'H' means "is happy", Q means "is quiet" and T means "is a theatre-goer", match the lists when the universe of discourse is people.
a)someone is not happy
b)someone happy is not quiet
c)not everyone is a theatre goer
d)no theatre goer is quiet
e)every theatre goer is both happy and quiet
f)if no theatre goer is happy, then some non theatre goers are quiet
g)not everyone who is happy is a theatre goer
h)there is at least one happy theatre goer who is quiet.
Note: let V=for all, ^=and, E=there exists, U=a U rotated clockwise half a turn (not even sure what it means.
1) ~VxTx
2) Ex~Hx
3) Ex((Hx ^ Tx) ^ Qx)
4)Vx(Tx U ~Hx) U Ex(~Tx ^ Qx)
5)Ex(Hx ^ ~Qx)
6)~Ex(Tx ^ Qx)
7)~Vx(Hx U Tx)
8)Vx(Tx U (Hx ^ Qx))
yea good luck, anything is appreciated.