This problem can be solved by trial and error (guess and check).
The analytic approach is to note that the general term in the sequence is:
an = 25-n
So we want 25-n < .01
We can take log (base...

You need the half angle formulas.
sin(x/2) = sqrt( (1-cos(x) ) /2 )
cos(x/2) = sqrt( (1 + cos(x) )/2)
There can be a ± in front of the square roots in some situations.

This problem can be solved very easily with calculus. At the maximum value of θ , the derivative of the
expression will be zero.
This derivative is 12 cos(θ) - 18 sin(θ) cos(θ) setting this to zero results in
12...

The polynomial remainder theorem states that if a polynomial , f(x), is divided by (x -a) using long division of polynomials, the remainder will be f(a). For example if the polynomial is
f(x) = x2 -9 and we divide by x-2, ...

The first statement is essentially true. The conduction band electrons in single domain of iron do have parallel spins. Since the electron has a magnetic moment, a magnetic field results.
The second is almost true. A voltage will be induced in a wire coil moving...

A special solution can be fairly easily guessed. It is
f = (1/(w -k)) cos(kx -wt) + C.
To this we would have to add a general solution to the homogenous equation df/dx + df/dt = 0
This general solution looks like ...

Well the sequence is
91 92 94 98
So the second term is 81.

The rational roots theorem says that the possible rational roots of this quartic are +1, -1, + 3, -3, +9, -9.
Trying all six gets just one hit g = -1
This means that the quartic is a product of (g+1) and a cubic polynomial. The cubic polynomial can...

Using standard trig identities, and multiplying through by sin a cos a, the starting equation can be
rewritten as
(sin a + cos a) (1 + sin a cos a ) = 7 sin a cos a - 1
Next, we replace sin a cos a ...

The standard form of the equation of a parabola is
f(x) = a x2 + b x + c.
With this standard from, the x coordinate of the vertex is -b/(2a).
Comparing with the function given, we see that
a = -9 ...

If A is a general square matrix, it can be written as
A = S + V where
S ≡ (AT + A) /2 and
V ≡ (-AT + A) /2
(T denotes transpose)
with these...

This problem can be set up as an optimization problem in calculus
The points on the parabolic curve can be written as (x, 11- x2)
so the upper right corner will have the coordinates of (x, 11-x2) for some optimal value of x
The width = 2x and the height...

This problem can also be solved using calculus techniques.
We need a figure similar to the one described in the non-calculus answer. If we call h the height above the ground at which the ladder touches the building, and x the distance from the foot of the ladder to the...

Rather than track your steps, I will show you how I would approach this problem.
Areas can be computed as integrals over either x or y. For the problem at hand, it is easier to think of x as function of y and to carry out the integral over y. Notice that the equations...

Power is normally voltage (RMS in the case of AC) times current (also RMS in the case of AC). So
P = 120 x 8 x .9 = 864 Watt
The 0.9 comes in because (again for AC) the voltage and current are not fully in phase.

Since the cdf is a smooth function for 0 < X < 1,
P( X < .8) is just cdf(.8) = .64

This problem can be solved by differentiation, with respect to time, of the equation
P V1.4 = C
The derivative of the right hand side is zero, and the derivative of the left hand side can be worked out
using the product rule and the power rule...

This integral is the evaluation of the integrand e^x^3 on the area in the x,y plane bounded by
the x axis, the line x = 1 and the curve y = x^2 . Integrals like this can written as either
∫ dy ∫ dx or ∫...

There are two ways do evaluate this volume.
The first is to use the high school formula for the volume of a pyramid. This is V = (1/3) base_area x height.
The base area (in the x y plane) is 1/2. The height = 3, so V = (1/3) (1/2) 3 ...

For a) the integral is ∫ dy ∫ dx 1 with the following limits:
The lower limit of the outer integral is 0
...