What is the smallest value of n such that an algorithm whose running time is 100n 2 runs faster than an algorithm whose running time is 2n on the same time.

n = 15

Step-by-step explanation:

For inputs of the value of n, the running time for the algorithm A is 100n^2 and that of B is 2^n.

If A is to run faster than B, 100n^2 must be smaller than 2^n.

Let's check from n = 1 to know the value of n that fits

n = 1

100(1)^2 > 2^1

100 > 2

n = 2

100(2)^2 > 2^2

400 > 4

n = 4

100(4)^2 > 2^4

1600 > 16

n = 8

100(8)^2 > 2^8

6400 > 2^8

n = 16

100(16)^2 < 2^16

25600 < 2^16

This implies that between n = 8 and 16, A starts to run faster than B

n = (8+16)/2 = 12

100(12)^2 > 2^12

14400 > 2^12

n = (12+16)/2 = 14

100(14)^2 > 2^14

19600 > 2^14

n = (14+16)/2

n = 15

100(15)^2 < 2^15

22500 < 2^15

At n= 15, A starts running faster than B

false

step-by-step explanation:

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation: