Write a function findEvenQueue that returns the first even number in the Queue. The function looks at the value at the front of the queue and checks if it’s even. If the value is even, then it returns that number. If the value is not even, then it removes that item from the queue and the process is repeated checking the front of the queue again. The function should return -1 if there are no even numbers. HINT: The program should make use of the peek method and the poll method of the queue.

You will be given two character arrays of the same size, one will contain a number of ships. ships will move around the character array based on which way they are facing and the route they are on. routes are given in the other array. the route consists of '-' and '|' for straight paths, '\' and '/' for curves, and '+' for intersections. there are ships on these routes. ships always face a direction, '^' for up, '> ' for right, 'v' for down, and '< ' for left. any time the ships hit a '\' or a '/' it will turn as you would expect a ship to turn (e.g. a '^' that moves into a '/' will turn right). at an intersection, ships will always continue straight through. all ships move at the same speed, ships take turns moving and all ships move during one 'tick'. the one in the most top left goes first, followed by those to its right, then the ones in the next row. it iterates along the rows and then down the columns. each ship moves one space on its turn moving along the route. your function needs to return the position of the first collision between two ships and the number of ticks before the crash occurred.

How much current flows through the alternator brushes? a. 2–5 a b. 25–35 a, depending on the vehicle c. 5–10 a d. 10–15 a

The initial tableau of a linear programming problem is given. use the simplex method to solve it. x 1 x 2 x 3 s 1 s 2 z 1 2 4 1 0 0 8 3 4 1 0 1 0 10 minus3 minus12 1 0 0 1 0 the maximum is nothing when x 1equals nothing, x 2equals nothing, x 3equals nothing, s 1equals3, and s 2equals0. (be sure to simplify to lowest terms if necessary.)