Answer:
attempts are required to find a matching pair if the digital fingerprint is 64 bits long.
attempts are required to find a matching pair if the digital fingerprint is 128 bits long.
Step-by-step explanation:
Each bit has two options. So
How many attempts are required to find a matching pair if the digital fingerprint is 64 bits long?
So for each of the 64 bits, we have the following number of options.
2 - 2 - 2 - 2 -... - 2
So, in all, there are
![T = 2^{64} = 1.85x10^{19}](https://tex.z-dn.net/?f=T%20%3D%202%5E%7B64%7D%20%3D%201.85x10%5E%7B19%7D)
options.
So,
attempts are required to find a matching pair if the digital fingerprint is 64 bits long.
128 bits long?
Using the same logic as the first question.
![T = 2^{128} = 3.40*10^{38}](https://tex.z-dn.net/?f=T%20%3D%202%5E%7B128%7D%20%3D%203.40%2A10%5E%7B38%7D)
So,
attempts are required to find a matching pair if the digital fingerprint is 128 bits long.
Answer:
x= 20
y = 60
Step-by-step explanation:
y = 3x
6(3x) = 360
3x = 360/6
3x = 60
x = 20
y = 3*20
y = 60
Answer:
x = -15
Step-by-step explanation:
-24 + 15 - 6 = x
-9 - 6 = x
-15 = x
Answer:
![n=-0.24](https://tex.z-dn.net/?f=n%3D-0.24)
Step-by-step explanation:
![3\left(-0.4n-0.7\right)=-3.2n-2.58](https://tex.z-dn.net/?f=3%5Cleft%28-0.4n-0.7%5Cright%29%3D-3.2n-2.58)
Use distributive property.
![-1.2n-2.1=-3.2n-2.58](https://tex.z-dn.net/?f=-1.2n-2.1%3D-3.2n-2.58)
![-1.2n-2.1+2.1=-3.2n-2.58+2.1](https://tex.z-dn.net/?f=-1.2n-2.1%2B2.1%3D-3.2n-2.58%2B2.1)
![-1.2n=-3.2n-0.48](https://tex.z-dn.net/?f=-1.2n%3D-3.2n-0.48)
![-1.2n+3.2n=-3.2n-0.48+3.2n](https://tex.z-dn.net/?f=-1.2n%2B3.2n%3D-3.2n-0.48%2B3.2n)
![2n=-0.48](https://tex.z-dn.net/?f=2n%3D-0.48)
![\frac{2n}{2}=\frac{-0.48}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2n%7D%7B2%7D%3D%5Cfrac%7B-0.48%7D%7B2%7D)
![n=-0.24](https://tex.z-dn.net/?f=n%3D-0.24)
Answer:
it will be 0
Step-by-step explanation:
x+y+2 = 0
x³+y³+2³ = 0
*8=2³