How many ways can david pick four of the first twelve positive integers such that no two of the numbers he picks are consecutive?

70 times

Step-by-step explanation:

5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 The smallest integer in our sequence is 2n = 2*17 = 34 Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

step- i d k and i d c : )

you: : 0

bye