Answer:
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Answer:
I was actually just looking to help with someone's schoolwork. . .
Explanation:
Since Wi-Fi Protected Access (WPA) fixes critical vulnerabilities in the earlier wired equivalent privacy (WEP) standard, the attacks that is related to encrypted wireless packets is option A: IV attacks.
<h3>Describe an IV attack.</h3>
A wireless network attack is called an initialization vector (IV) attack. During transmission, it alters the IV of an encrypted wireless packet. One packet's plaintext can be utilized by an attacker to calculate the RC4 key stream produced by the IV employed.
Note that A binary vector used to initialize the encryption process for a plaintext block sequence in order to boost security by adding more cryptographic variance and to synchronize cryptographic hardware. The initialization vector is not required to be kept secret.
Learn more about Wi-Fi Protected Access from
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See full question below
Wi-Fi Protected Access (WPA) fixes critical vulnerabilities in the earlier wired equivalent privacy (WEP) standard. Understanding that WPA uses a combination of an RC4 stream cipher and Temporal Key Integrity Protocol (TKIP), this makes a wireless access point NOT vulnerable to which of the following attacks when related to encrypted wireless packets?
IV attacks
Malware
Ransoware
Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
Answer:
Runtime error probably. The program won't make it past the while loop in the code.
