Alice and Bob agree to communicate privately via e-mail using a scheme based on RC4, but they want to avoid using a new secret key for each transmission. Alice and Bob privately agree on a 128-bit key k. To encrypt a message m consisting of a string of bits, the following procedure is used. 1. Choose a random 80-bit value v 2. Generate the ciphertext c = RC4(v || k) ⊕ m 3. Send the bit string (v || c) Suppose Alice uses this procedure to send a message m to Bob. Describe how Bob can recover the message m from (v || c) using k.

You need to design a circuit that implements the functions in the following table: s0 s1 function0 0 a + 10 1 a – b1 0 a + b1 1 a – 1s0 and s1 are 1-bit control inputs to select the function of the circuit. inputs a and b are 4-bitnumbers in 2s complement form. the output is also a 4-bit number in 2s complement form. you are allowed to use only one ttl 7483 4-bit adder to implement all the functions. but anynumber of other components (except the adder) can be used. hint: design a combinational logic circuit to modify the input b and the “carry input” of theadder depending on the control inputs s0 and s1.important: lab grade will depend on the working of the circuit & will be checked of by your labinstructor.1. is the output valid for the following input combinations: a. s0 = 0, s1 = 0, a = 7, b = 3? b. s0 = 0, s1 = 1, a = 7, b = 3? c. s0 = 1, s1 = 0, a = -4, b = -5? d. s0 = 1, s1 = 1, a = -8, b = 6? 2. what is the range of inputs (for both a and b) that will produce the valid output for all the functions?

Me dangers of social media and the internetexplain what each means: 1) social media and phones have become an addiction.2) outside people have access to you all the time.3) cyberstalking4) cyberbullying5) catphishing6) viruses7) identity theft8) credit card fraud9) hacking10) money schemes

Use a software program or a graphing utility with matrix capabilities to solve the system of linear equations using an inverse matrix. x1 + 2x2 − x3 + 3x4 − x5 = 6 x1 − 3x2 + x3 + 2x4 − x5 = −6 2x1 + x2 + x3 − 3x4 + x5 = 3 x1 − x2 + 2x3 + x4 − x5 = −3 2x1 + x2 − x3 + 2x4 + x5 = 5