Assume that a and b share a long-term 256-bit aes key k. our key exchange protocol has two goals: (1) let a and b agree on a new fresh session key ks to be used for a current session, and (2) perform mutual authentication. nb is a nonce and ks is a new session key generated by b. assume that ks and nb are of the same bit-length and both are always chosen at random. denotes aes encryption of everything inside () with key k. denotes sha2-based hmac of everything inside () using key k. protocol: 1. a > b: ek(a, b, "hello") what (if any) are the vulnerabilities of this protocol? and, how can you fix them? again, keep your answers limited to 8-10 lines.

hi, i am not sure about this one but i think its d,

hope this !

step-by-step explanation:

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.