answer:If the final length of the movie is 57 minutes after cutting a 3-minute scene, then the original length of the movie would be: 57 minutes (final length) + 3 minutes (cut scene) = 60 minutes (original length) So, the original length of the movie was boxed{60} minutes.

answer:To start, the condition that vector overrightarrow {a} is perpendicular to vector overrightarrow {b} implies that their dot product is zero. We have the following vectors: overrightarrow {a}=(4, 2), overrightarrow {b}=(x, 3). The dot product overrightarrow {a} cdot overrightarrow {b} is calculated as: overrightarrow {a} cdot overrightarrow {b} = 4x + 2 cdot 3. For the vectors to be perpendicular, the dot product must be zero: 4x + 6 = 0. Solving for x, we subtract (6) from both sides: 4x = -6. Divide by (4) to isolate (x): x = -frac{6}{4}. Simplify the fraction: x = -frac{3}{2}. Therefore, the value of x is (boxed{-frac{3}{2}}).

answer:To find the volume of a trapezoidal prism, we first need to find the area of the trapezoidal base and then multiply it by the height of the prism. The area ( A ) of a trapezium (trapezoid) can be found using the formula: [ A = frac{1}{2} times (a + b) times h ] where ( a ) and ( b ) are the lengths of the parallel sides, and ( h ) is the distance (height) between the parallel sides. In this case, the parallel sides of the trapezium are 20 cm and 15 cm, and the distance between them is 14 cm. So we have: [ A = frac{1}{2} times (20 + 15) times 14 ] [ A = frac{1}{2} times 35 times 14 ] [ A = 17.5 times 14 ] [ A = 245 text{ cm}^2 ] Now that we have the area of the trapezoidal base, we can find the volume ( V ) of the prism by multiplying the area of the base by the height of the prism: [ V = A times text{height of prism} ] [ V = 245 text{ cm}^2 times 10 text{ cm} ] [ V = 2450 text{ cm}^3 ] So the volume of the trapezoidal prism is boxed{2450} cubic centimeters.

answer:First, factor the denominator similar to the strategy used in the original problem: [1 + 3^n + 3^{n + 1} + 3^{2n + 1} = (1 + 3^n) + 3^{n + 1} (1 + 3^n) = (1 + 3^n)(1 + 3^{n + 1}).] Next, rewrite the numerator using the factored form: [3^n = (1 + 3^{n + 1}) - (1 + 3^n) = 3^n,] thus, [frac{3^n}{1 + 3^n + 3^{n + 1} + 3^{2n + 1}} = frac{(1 + 3^{n + 1}) - (1 + 3^n)}{(1 + 3^n)(1 + 3^{n + 1})} = frac{1}{1 + 3^n} - frac{1}{1 + 3^{n + 1}}.] Now, calculate the sum: begin{align*} sum_{n = 1}^infty frac{3^n}{1 + 3^n + 3^{n + 1} + 3^{2n + 1}} &= left( frac{1}{1 + 3} - frac{1}{1 + 3^2} right) + left( frac{1}{1 + 3^2} - frac{1}{1 + 3^3} right) + left( frac{1}{1 + 3^3} - frac{1}{1 + 3^4} right) + dotsb &= frac{1}{4} - frac{1}{1 + 3^{n+1}} text{ as } n to infty. end{align*} Since frac{1}{1 + 3^{n+1}} to 0 as n to infty, the series converges to: [boxed{frac{1}{4}}.]