ABC is a triangle where AN is perpendicular to CB and BM perpendicular to AC. The length of BC is 10, that of AC is 12 and that of AN is 8. Find the length of BM.
The area A of the given triangle may be calculated using the two altitutdes as follows A = (1/2)(AN)(BC) or A = (1/2)(BM)(AC) Hence (1/2)(AN)(BC) = (1/2)(BM)(AC) Multiply both sides by 2 and substitute known lengths 8 * 10 = MB * 12 Multiply both sides by 2 and substitute known lengths BM = 80 / 12 = 6.7 (approximated to one decimal place)
Use the Intermediate Value Theorem to show that the function f(x) = x^3 + x^2 + x +1 takes on the value 12345 at least once in its domain.
Since f is a polynomial, it is continuous over all real numbers. f(0) = 0 < 12345 and f(12345) = 123453^3 + 123452^2 + 12345 + 1 > 12345 (since all terms are positive). So by the IVT, f(c) = 12345 for some c between 0 and 12345
Example 1: Solve and find a general solution to the differential equation. 2 y ' = sin(2x)
Solution to Example 1: Write the differential equation of the form y ' = f(x). y ' = (1/2) sin(2x) Integrate both sides ò y ' dx = ò (1/2) sin(2x) dx Let u = 2x so that du = 2 dx, the right side becomes y = ò (1/4) sin(u) du Which gives. y = (-1/4) cos(u) = (-1/4) cos (2x)