Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
Hello <span>Areyano7475
</span>
Question: T<span>he term drive app is used to describe applications stored on a computer true or false
Answer: False
Hope this helps
-Chris</span>
Answer:
in my opinion 4
Explanation:
when the system is available to users
(sorry and thanks)
Answer:
I am doing it with python.
Explanation:
nums = '9 -8 -7 -6 -5 -4 -2 0 1 5 9 6 7 4'
myfile = open('data.txt', 'w')
myfile.write(nums)
myfile.close()
myfile = open('data.txt', 'r')
num1 = (myfile.read())
num1 = num1.split()
print(num1)
print(type(num1))
for x in num1:
x = int(x)
if x < 0:
minus = open('dataminus.txt', 'a')
minus.write(str(x) + ' ')
minus.close()
elif x>= 0:
plus = open('dataplus.txt', 'a')
plus.write(str(x)+' ')
plus.close()
