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:
Given equation:
Take natural logs of both sides:
Expand brackets:
Collect like terms:
Factor left sides:
Simplify:
Answer:
Statement 1 & Statement 3
Step-by-step explanation:
4 ÷ 10/3
4 × 3/10
12/10
6/5
f(x + 1) = (6/5) × f(x)
f(x) = (10/3) × (6/5)^(x-1)
f(x) = (10/3) × (5/6) × (6/5)^x
f(x) = (25/9) × (6/5)^x
The domain is all natural numbers and the range is real numbers
Let x be the first number:
1st number: = x
2nd consecutive multiple of 6 = x+6
3rd consecutive multiple of 6 = x+12
4rth consecutive multiple of 6 = x+18
Their sum = 156 → x+(x+6)+(x+12)+(x+18) = 156
4x +30 = 156
4x = 120 and x = 30
The numbers are: 30,36,42,48
