This is an expression, not an equation. However, the expression can be expressed as a partial fraction decomposition. The denominator factors as ( 2x -1) (x + 2) (x -2). {Factoring by groups works here} This...
This is an expression, not an equation. However, the expression can be expressed as a partial fraction decomposition. The denominator factors as ( 2x -1) (x + 2) (x -2). {Factoring by groups works here} This...
This problem is a typical optimization problem. There is a cost function , C(x,z) and a constraint condition: 50 = x2 z C(x,z) = 2 x2 5 + 4 xz 2 The 2 is there...
The triple integral is M = ∫dx ∫dy ∫dz k z Since the boundary condition on x and y is the circle of radius a, the easiest way to proceed is to change ∫ dx ∫ dy to ∫ρ dρ ∫ dθ That is - change to polar coordinates. ...
The first thing to note is that the second compound should be NaOH. The OH group (hydroxide) stays together, so you should keep track of it as a unit, not as separate O and H. You need a 3 in front of the NaOH to balance up the OH groups (three on each side)....
The average rate of change is similar to an average velocity in physics related problems. The average velocity is the change in displacement divided by the time that the change took. So by analogy: The average rate is Ravg = ...
The objective function, Q, is the potential profit that could be made. From the problem statement Q(x) = p(x) x - 1 x = (8 - ln(x)) x -x = 7x - x ln(x) The calculus approach is to find the value of x that maximizes Q. ...
Work, W, is Power, P, multiplied by time, t. The work done by the engine in time t is thus W = P t. The work energy principle is that the work done on a system is equal to the change in energy, ΔE, of the system. Since there...
The problem is symmetric about the y axis, so one of the lines will have positive slope with point of tangency on the right side of the parabola, while the other will have negative slope with point of tangency on the left. The equation of the first line will be of the form y...
The answer depends on the mass of the bowling ball and on how elastic the collision actually is. If the the collision is assumed to be an elastic head-on collision, both energy and momentum are conserved. Head- on collisions are inherently one dimensional. For these,...
The first step is to write the two balanced combustion equations: CH4 + 2 O2 ==> CO2 + 2 H2O C2H6 + (7/2) O2 ==> 2 CO2 + 3 H2O Let the...
Multiply both sides by exp(-x) exp(-y) to get exp(-y) dy = - exp(-x) dx Then integrate both sides to get exp(-y) = - exp(-x) -C This can be rearranged as ...
You are on the right track. r sin(θ) = r cos(θ) then divide both sides by r to get sin(θ) = cos(θ) the solution to this is θ = pi/4 ( 45 °) Straight lines passing...
The four sublevels used in electron configurations are s, p, d, f An s sublevel can hold a maximum of 2 electrons a p sublevel can hold a maximum of 6 electrons a d sublevel can hold a maximum of 10 electrons an f sublevel can...
Using the TI-84 calculator, the inverse of the coefficient matrix is calculated to be the matrix 1 5 13 0 -1 -3 1 6 15 Since the inverse exists, the solution to the system is obtained by multiplying the column...
The energy of one quantum of light (a photon) is given by E = h c / λ where h is Planck's constant, c is the speed of light and λ is the wavelength. Using the standard values of h and c and λ = 8.9 E-11 m, ...
The molar mass of OH- is 17 g. So a concentration of 0.3 g/L of OH- is .3/ 17 = 0.0176 molar. The pOH is then pOH = log(.0176) = 1.75. The corresponding pH is ...
It appears that the expression nt8 just before the equals sign should be n+8. With that the equation reads: n3 - 8n2 -n + 8 =0. The factor by grouping approach works. Rewrite as n2 (n...
Since the function is f(x) = 2 ex +x, the derivative , f' , is 2 ex +1. The derivative is the slope, so that at the point (0,2), the slope (m) is 2 e0 + 1 = 3. Once...
The approach to limiting reactant problems with two reactants is to compute the mass of product formed two ways: 1) by assuming that the first reactant is limiting and 2) by assuming that the second reactant is limiting. ...
The portion of range [pi, 3pi/2) of the arcsec corresponds to the third quadrant of the sec function. In this quadrant the sec is negative, so arcsec(-2) can be an angle in the third quadrant. To find the angle one wriites ...