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Finding W T In Second Order Repeated Roots

What is a two dimensional figure with non congruent side?

Like... a scalene triangle (a triangle with sides of 3 different lengths) is a 2d figure with non congruent sides.

How can I find a path of a given length between two nodes in a weighted (undirected) graph?

If your desired length is small, then the color-coding method is pretty reasonable. See Color-coding. If k is the length of your desired path, the runtime of the algorithm is [math]O(k^k n)[/math].The algorithm consists of many iterations where you randomly color all of the vertices with [math]k[/math] colors independently with colors 1, 2, ..., k. In the resulting colored graph, you then look for a path of length [math]k[/math] from your starting vertex [math]u[/math] to your terminal vertex [math]v[/math] that contains vertices colored with 1, 2, 3, ..., k. This can be found using a BFS that, when at a vertex with color i, only proceeds to vertices with color i + 1. With probability 1/k^k, a path of length k, if it exists, will be colored with colors 1, 2, ..., k in order. Therefore, you need around k^k iterations to discover the path with high probability. If your graph has many paths of this length between u and v, it will need fewer iterations to find the path.

What is the difference between linear and quadratic equations?

Linear equations are generally in this format:2x - 4 = yAnd when graphing linear equations: (PS the equation and the graph below do not match - just a general format of linear equations’ graph)Whereas quadratic equations are in this type of format:x^2 + 5x + 6 = yAnd when graphing quadratic equations: (again, the equation and graph do not match)***So basically, when we say linear equations, we mean those equations that have a maximum of only one x-intercept, or in other words, when y = 0, they only have a maximum of one coordinate that passes through the x-axis -> as seen in the graph above.However, quadratic equations are a little bit different. Whereas linear equations have only a maximum of one x-intercept, quadratic equations’ graph (a parabola) has a maximum of two x-intercepts, meaning that when y = 0, the parabola passes through a maximum of two coordinates in the x-axis -> as seen in the graph above.PS- A parabola and a linear equation graph don’t have any minimum of points: for ex: a linear equation graph can have 0 x-intercepts, and a parabola can have 1 or even no x-intercept at all (Just some general information that I thought you might want to know for greater understanding).

Identify which property of real numbers the statement illustrates.?

Property of Multiplication.

If a pendulum clock is taken to the moon, will it gain or lose time?

A simple pendulum oscillates at a period given by the expressionP = 2 pi sqrt(L/g),where L is the length of the pendulum and g is the acceleration of gravity. The mass of the pendulum bob does not enter. On the Moon the acceleration of gravity is less than it is on the surface of the Earth, so the period on the Moon will be greater than it would be at Earth’s surface. This results in fewer “ticks” of the clock in a given period of time. Therefore a clock on the Moon will run slower than an identical one on Earth. The Moon clock will lose time compared to the Earth clock.

What is the formula for the total kinetic energy of a mole of ideal gas molecules?

kT is equal to 2/3 of molecule kinetic energy , so 3/2kT is average molecular energy now if you need to find what energy 1mole has you just need to multiply this number by Avogadro’s number Na=6,022×10^23

A ball is dropped from a height of 45m. What will be the time to reach the ground?

The distance is 45m. (s)Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)We have to find the time (t)By the 3rd equation of motion,s = ut + 1/2 at^2 (here u is the initial velocity of the ball)Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.(substituting the values in the formula)45 = 1/2 x 9.8 x t^245 = 9.8/2 x t^245 = 4.9 x t^2 (9.8/2 = 4.9)(divide 4.9 on both sides)45/4.9 = 4.9 x t^2 / 4.99.1836 = t^23.030 = tSo it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).

Solve the differential equation using classical method and the given initial conditions?

solve the differential equation using classical method and the given initial conditions

[ d^2x/dt^2 ] + [ 2 dx/dt ] + [ x ] = 5 e^-2t +t

x (0) = 2; dx/dt(0) = 1

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