There is a straightforward calculus approach to this problem. Since 6,8,10 are the side lengths of a right triangle (double the 3,4,5 triangle) , there is probably a purely geometric approach as well.
The calculus approach is to note that the area bounded by a chord (length...

The best approach seems to be to spit into the sum of two integrals
Integral = ∫dx 2/sqrt(1- 25 x2 ) + ∫ dx 5 x / sqrt( 1- 25 x2)
Both of these integrals can be done by substitution. For the first one use a trig...

The implicit assumption here is that the bulb lifetime is normally distributed. The fraction of bulbs that burn out within 11000 hours is the area under the relevant normal curve between - ∞ and 11000.
The easiest way to get a numerical answer is to use a graphing...

The determinant of the coefficient matrix ( the 2 x2 matrix formed from the first two columns) is zero. This means that the two underling linear equations are not independent.
If h is chosen as -2, the two equations would be ...

A simple example is the set { 1, 2, 3,4, 5,6,41} The median is 4, the mean is 62/7 ~ 8.86 which is greater than 4 , the range is 40. There is no mode because all members are unique. The mean is ~ 8.86, the standard deviation is ~ 13.2, ...

A trick answer to this is that all seven number are zero. Zero is a whole number, the mean and median are both zero and the conditions on the greatest and least in terms of the median are satisfied. I looked for other solutions but did not find any.
...

The difference between the mean score (74) and a score of 99 is 25. Dividing by the standard deviation (8), gives 3.125 . This means that a score of 99 is more than three standard deviations above the mean. Clearly, there will be few scores more than 99. ...

For part a) To locate the zero of the electric field it is useful to consider three intervals of x:
(-∞ , -5) ; (-5, 0) ; (0, ∞)
For the first interval, the contribution from the positive charge...

Since the electric field is 8,25 x 104 V/m and the plate separation is 2.15 x 10-3 m, the electric potential across the device is the product of these ; V = 177.375 V. Since Q = V C, C = ...

The reason why the double filters reach a higher terminal velocity than a single one is the same as the reason that a 120 kg parachutist falls faster (terminal speed) than does a 60 kg parachutist. (Identical parachutes assumed).
At terminal speed, the air drag force...

I assume that the curve is y = x2 (a parabola)
This is an optimization problem. The task is to minimize the distance, d, between a point on the parabola and the point (3,0). It practice, it is somewhat easier to minimize...

This is an interesting functional form, but the standard approach still works.
The analytic form of the derivative dy/dx can be found by the method of implicit differentiation. The result is
dy/dx = - (x/y)1/3 . Evaluating...

The augmented matrix is the 3 x 3 coefficient matrix with a fourth column added.
{ 7 4 7 74
3 7 5 51
2 2 9 71 ...

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...