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Count Iblis
Questions (11)
Original qustion posted here:
http://www.jiskha.com/display.cgi?id=1239379999 If you do computations Modulo some number, say
0 answers
709 views
Let's denote the three numbers by a1, a2, and a3. Consider the third degree polynomial:
p(x) = (1 + a1 x)(1 + a2 x)(1 + a3 x)
0 answers
668 views
momentum
What is a kilogram meter per second? vesc=(2GM/R)1/2
0 answers
366 views
This is a reply to the question posted here
http://www.jiskha.com/display.cgi?id=1178989522 As I explained there, you can find
0 answers
1,070 views
This reply to some student got deleted, so I'm reposting it.
Cos(pi/4 + h) = Cos(pi/4)Cos(h) - Sin(pi/4)Sin(h) = 1/sqrt(2)[1 - h
16 answers
723 views
correction:
9/25(u+7)u^11 du = 9/25(u^12 + 7 u^11) du and you can easily integrate this indefinate integral of 9x(5x-7)^11 dx
0 answers
485 views
|2x + 4| is always larger than or equal to zero. This means that the left hand side of the equation is always 6 or less, so
0 answers
1,190 views
Ideal Gas Law:
PV = N k T ---> N = PV/(k T) N = 1.387*10^(29) mass of one helium atom is approximately 4*u = 6.642*10^27 kg 4u*N
0 answers
632 views
i^6 = (i^2)^3 = (-1)^3 = -1
The square, not the square root of -1 is 1. :) B.t.w., can you prove that -1 times -1 is 1? Hint, try
2 answers
839 views
RP^2=RQ^2 + QP^2 -2(RQ)*(QP)cosQ (1)
QP = 24 (km/min) t (2) Diffentiate both sides w.r.t. the time t: 2 RP* d Rp/dt = 2 t [24
2 answers
1,000 views
Hi Jen,
I've seen the drawing. My site is down at the moment. Draw a line from Venus (either of the two positions) to the line
0 answers
1,085 views
Answers (884)
You can use generating functions. Consider the function f(y) = sum over x1,...x5 of y^(x1 + x2 + x3 + 2x4 +x5) where the summation over x1,...,x5 is from 0 to infinity. This then factors into geometric series, the result is: f(y) = 1/(1-y)^4 1/(1-y^2) =
sin^2(theta) = 1-cos^2(theta) Therefore: sin^8(theta) = [1-cos^2(theta)]^4 = 1 - 4 cos^2(theta) + 6 cos^4(theta) - 4 cos^6(theta) + cos^8(theta)
If a number X is divisible by Y, then the remainder of X after division by Y is zero. Calculating the remainder after division can be greatly simplified, you often don't need to actually divide X by Y to calculate the remainder. The remainder after
The fraction is: p(x)/(x-5)^3 with p(x) = 3x^2-34x+97 Then let's expand p(x) in powers of (x-5). If we put x = 5+t then we have: p(5+t) = 2 - 4 t + 3 t^2 Therefore, partial fraction expansion is: 2/(x-5)^3 - 4/(x-5)^2 + 3/(x-5)
Put T = 1 hour. Then in the rest frame of an observer on the ground, the spacecraft will travel distance of v T, during a time of T, this means that the space-time interval between two points on the trajectory a time T apart is given (in c = 1 units)by s^2
p = gamma(v) m v where gamma(v) = 1/sqrt[1-v^2/c^2]
Do a partial integration twice, e.g. integrate the exponential function, and then in the integral of e^(5 t) sin(5 t), you again integrate the e^( 5 t) factor (and not the sin(5t) ). You then get the the same integral back but with additional terms. You
You just look at the expression and then try defining a function u(x) such that in terms of u the integral will simplify. There is no "wrong choice" because the integral in terms of u will always be the same as the original one. However, the goal is to
We have: |+> = 1/sqrt(2)[|0> + |1>] |-> = 1/sqrt(2)[|0> - |1>] M = 3|+>
The bulk modulus K used here is the so-called adiabatic bulk modulus and this is related to the pressure P according to: K = gamma P where gamma = cp/cv = approximately 1.4 for air. So, you know the pressure and the density and then you can find the
Assuming the lottery is fair, all combinations of numbers have equal probability. The only thing to consider when choosing a set of numbers is then to choose a set that no one else is likely to choose. Then when you do win, you won't end up having to share
Good old Pythagoras's theorem says that: ds^2 = dx^2 + dy^2 We can write: dx = [-5 sin(t) + 5 sin(5t)] dt dy = [5 cos(t) - 5 cos(t)] dt ds^2 = dx^2 + dy^2 = (using sin^2(x) + cos^2(x) = 1 twice for x = t and x = 5t) 25 [2 - 2 sin(t) sin(5t) - 2 cos(t)
The force on the ball is unbalanced, the floor exerts a force on the ball which causes the ball to accelerate away from the floor. However, this doesn't explain why the ball changes its shape. If you hold the ball in your hands and compress it, it will
g'(x)=af'(x). Try to prove this using the definition of the derivative.
At a fixed angle from the slits, if you decrease the distance between the slits the path length difference decreases. So, the angle needs to increase to get back to the same path length difference (half the wavelenght for a fringe).
Method 1: Assign a different number to each student ranging from 1 to 90, numbers 1 to 30 go in group 1, 31 to 60 go to group 2 61 to 90 go to group 3. All possible partitions are obtained with equal probablity by a random assignment if these numbers, it
Our time and energy used by our computers and the internet servers.
The probablity of any specific sequence of heads and tails is (1/2)^4, because the probability of getting a head at each throw is 1/2. So, the probability of getting exactly 2 heads is 1/16 times the number of ways of having two heads in a sequence of 4
You get a contradiction, i.e. a false statement. This is because you are writing down a set of equation, which already supposes that there does exist at least one solution. So, if there are no solutions that supposition is false, so you must get a
int y dy/(y^2-2y-3) = int 1/2 2y dy/(y^2-2y-3) = int 1/2 (2y - 2 + 2) dy/(y^2-2y-3) = 1/2 Log(y^2 - 2 y - 3) + int dy/(y^2-2y-3) int dy/(y^2-2y-3) = int dy/[(y-3)(y+1)] 1/[(y-3)(y+1)] = 1/4 [1/(y-3) - 1/(y+1)] -----> int dy/(y^2-2y-3) = 1/4
(4n-7)/(4n+9) = [(4n+9) -16]/(4n+9) = 1 - 16/(4n+9) The limit thus exists and is equal to 1. Clearly, for every epsilon > 0, there exists an N such that A_n will be within epsilon of the limiting value for all n > N, taking N = 4/epsilon will do.
|a_n| = |sin(2/n)| Since |sin(x)| N the absolute value of the difference of a_n and the limit is less than epsilon.
If f(x1,x2) is a function of x1 and x2 which have independent uncertainties sigma1 and sigma2, then the uncertainty in f(x1,x2) is: sigmaf = sqrt[sigmaf1^2 +sigmaf2^2] where sigmaf1 and sigmaf2 are the contributions to the uncertainty in f coming from x1
In the co-rotating frame, define an angular variable alpha such that the channel is at alpha = 0. You can enforce that the ball is at alpha = 0 using a Langrange multiplier, so you treat the alpha coordinate as a dynamic variable for the ball. The kinetic
Each card has equal probability to end uop in the hand of the different players. So, for each ace you have 4 equaly likely choices for the players they will end up at, there are thus 4^4 = 2^8 ways the aces can end up in the hands of the players. There are
http://www.columbia.edu/~vjd1/carbon.htm "some examples: If CO2 concentration increases in the atmosphere because of an increased rate of outgassing, global temperature will rise. Rising temperature and more dissolved CO2 will lead to increased weathering
sin(x) = sin[2(x/2)] = 2 sin(x/2) cos(x/2) Draw a right triangle with one angle equal to x/2. If you make the length of the side opposite to that angle equal to t = tan(x/2) then the length of the side side orthogonal to it that connects to that angle will
Without the hole the field would be sigma/epsilon_0 directed in the radial direction. Then instead of cutting a hole, you can put a surface charge of -sigma there. The field is then sigma/epsilon_0 plus the field due to the additional surface charge
sqrt(x^2+6x+3) = x sqrt(1 + 6/x + 3/x^2) Using the series expansion: (1+y)^p = 1 + p y + p (p-1)/2 y^2 + ... for p = 1/2 gives: sqrt(1 + 6/x + 3/x^2) = 1 + 3/x + O(1/x^2) Where the O(1/x^2) means that there exists an R and a constant c such that for x
a) If you take two polynomials p1(t) and p2(t) that have a constant term 1 and take the linear combination a p1(t) + b p2(t) then what is the constant term of that linear combination? For arbitrary a and b does this then belong to the set of all the 4th
The secoind derivative of a product of two functions f(x) and g(x) is: f''(x) g(x) + 2 f'(x) g'(x) + f(x)g''(x) If you take f(x) = x and g(x) = arcsin(x) then the first term is zero, so you only have to evaluate the last two terms. If you add them up you
http://www.jiskha.com/display.cgi?id=1382331843
The charge between radii r and r + dr on the disk is: sigma 2 pi r dr The contribution to the potential from this charge is: sigma/(4 pi epsilon) 2 pi r dr/sqrt(r^2 + x^2) Integrating over r from 0 to R gives: V(x) = sigma/(2 epsilon) [sqrt(R^2 + x^2) - x]
None of them. Unless ψ1 and ψ2 are solutions with the same energy eigenvalue (in which case we say that this eigenvalue is degenerate), the linear combination aψ1+bψ2 won't be a solution to the time independent Schrödinger equation.
The Klein-Gordon equation.
Determine on what intervals you have that: |x^2 + x | = x^2 + x and where do you have that: |x^2 + x | = -(x^2 + x) You can then compute the derivative on these intervals and see if what you get is the same as |2x + 1| for all x.
The cloud would be able to rise higher were it not for the fact that the troposphere ends at a certain height in the atmosphere (between 10 to 20 km altitude, at the equator it's 20 km near the poles it's 10 km). What happens is that the temperature
See here for the correct solution: http://www.jiskha.com/display.cgi?id=1381448688
Hadamard transform is defined as: U|0> = 1/sqrt(2) [|0> + |1>] U|1> = 1/sqrt(2) [|0> - |1>] The state is: 1/sqrt(2)|00> + e^iphi/sqrt(2)|10> We then have that: = 0 You can evaluate the l.h.s. by letting U act on the bra vector. Since U equals its own
Hint: What is g(0)?
m = 938.3 MeV/c^2 v = 7500 km/s m v = 938.3 MeV/c^2 7500 km/s = 938.3 MeV/c (7500 km/s)/c = 23.47 MeV/c
This is 1 - probability that each person will stay in a different hotel. There are 6!/2! ways to assign hotels to the four persons such that each person stays in a different hotel. The total number of ways to assign hotels to the person without any
Correction of the last part: The kinetic energy at the start of the rough surface is thus: E = E1 [cos(theta) - mu_1 sin(theta)]^2 The distance the block will slide is thus given by: d = E/(mu_2 m g) = L/mu_2 [cos(theta) - mu_1 sin(theta)]^2* [sin(theta) -
The height is L sin(theta). The decrease in the gravitational potential energy is thus m g L sin(theta). The magnitude of the component of the gravitational force orthogonal to the incline is m g cos(theta), therefore the normal force is equal to m g
This amounts to coloring 9 balls with 4 colors. We can represent a coloring as a string of 9 o's and 3 |'s. E.g. 000|00|000|0 represents 3 balls with color 1, 2 balls with color 2, 3 balls with color 3 and 1 ball with color 4. The number of different
http://www.jiskha.com/display.cgi?id=1381249499
The tension in a point of the rope is defined as follows. It is the magnitude of force that the part of the rope on one side of the point exerts on the part of the rope on the other side of the point. The direction of the force then depends on which sides
It's the number of ways you can put precisely 3 girls next to each other plus the number of ways you can put preciesly 4 next to each other. To get precisely 3 girls next to each ther, you can choose the 3 girls and the order in which they will take their
Put v = 19 m/s and alpha = 41°. v cos(alpha) is the horizontal speed of the ball and v sin(alpha) is the vertical speed of the ball. The equations for the vertical and horizontal motion of the ball are independent of each other. This allows you to
http://www.random.org/ 0.1526961 -0.4481168 -0.6982344 -0.6276160
