Answer:
The three types of user mode to the kernel mode transferred occurred due to the:
- It is mainly occurred due to the interrupt when, it send to the central processing unit (CPU).
- It also occurs due to the hardware exception and when the memory is access illegally as it is divided by the zero.
- It is mainly implemented or executed by the trap instruction as the system are basically executed by the program.
Answer / Explanation:
195.200.0.0/16
Note: Class C address can not be assigned a subnet mask of /16 because class c address has 24 bits assigned for network part.
2ⁿ = number of subnets
where n is additional bits borrowed from the host portion.
2ˣ - 2 = number of hosts
where x represent bits for the host portion.
Assuming we have 195.200.0.0/25
In the last octet, we have one bit for the network
number of subnets = 2¹ =2 network addresses
number of host = 2⁷ - 2= 126 network addresses per subnets
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 A. Hope it helped c:
