Answer:
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
Answer:
x = 
and x = 
Step-by-step explanation:
Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: 
or 
.
Here, a = 6, b = -24, and c = 1. Plug these in:
AND
Thus, the zeroes are: x = and x = .
I’m assuming it would just be x+3
The answer is D. You’d need to minus the area of the circle away from the area of the triangle to get the area of the shaded region
