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)

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step-by-step explanation:

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guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

a

step-by-step explanation: