Master theorem: t(n) = 3t(n/3) + o(log n)? hello, i've been trying to solve the bounds for the recurrence stated in the title: i know a = 3, b = 3, and f(n) = o(log n), n^log(a) of base b --> n^log(3) of base 3 --> n^1 = n. so i get θ(n). now i proceed to use case 1 of master theorem: if f(n) = o(n^log(a-e) of base b) for some constant e > 0, then t(n) = θ(n^log(a) of base b) since f(n) < n^log(a) of base b, asymptotically. i believe the solution should be θ(n) but where i am stuck is proving that f(n) is polynomially smaller, when using case 1, the ratio of f(n) / n^log(a) of base b: = log n / n, so that the solution θ(n) is true any with this is appreciated

Drag and drop the answers into the boxes to complete this informal argument explaining how to derive the formula for the volume of a cone. since the volume of a cone is part of the volume of a cylinder with the same base and height, find the volume of a cylinder first. the base of a cylinder is a circle. the area of the base of a cylinder is , where r represents the radius. the volume of a cylinder can be described as slices of the base stacked upon each other. so, the volume of the cylinder can be found by multiplying the area of the circle by the height h of the cylinder. the volume of a cone is of the volume of a cylinder. therefore, the formula for the volume of a cone is 1/3 1/2 1/3πr^2h 1/2πr^2h πr^2h πr^2

Solve the system by the elimination method. x + y - 6 = 0 x - y - 8 = 0 when you eliminate y , what is the resulting equation? 2x = -14 2x = 14 -2x = 14

What were mkh company's cash flows from (for) operating activities in 20x1? $(180,300) $233,100 $268,200 $279,400?