Which shows a recursive and an explicit formula for a sequence whose initial term is 14 and whose common difference is −4? a. a(1) = 14; a(n) = (n − 1) −4; a(n) = 14 + (n − 1)(−4) b. a(1) = 14; a(n) = (n + 1) −4; a(n) = 14 + (n − 1)(−4) c. a(1) = 14; a(n) = (n − 1) − 14; a(n) = 14 + (n − 1)(−14) d. a(1) = 14; a(n) = (n + 1) − 14; a(n) = 14 + (n + 1)(14)

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

2x^2+5x+17

your welcome. : )

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8

step by step explanation: when u removing the parentheses and combined them that would be : x+3y-x-y=2y