Answer:
x = int(input("What grade are you in? "))
if(x == 9):
print("Freshman")
elif(x == 10):
print("Sophomore")
elif(x == 11):
print("Junior")
elif(x == 12):
print("Senior")
else:
print("Not in High School")
Explanation:
Answer:
i). Signed magnitude
Five bit representation = 11111
For positive 5 bit representation = 01111 = +15
For negative 5 bit representation = 11111 = -15
ii). One's complement
For positive 5 bit representation = 01111 =+15
For negative 5 bit representation = 10000 = -15
iii). Two's compliment
For positive 5 bit representation = 01111 = -15
For negative 5 bit representation = 10001 = +15
Explanation:
User Devices. Users of wireless LANs operate a multitude of devices, such as PCs, laptops, and PDAs. ...
Radio NICs. A major part of a wireless LAN includes a radio NIC that operates within the computer device and provides wireless connectivity.
or routers, repeaters, and access points
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:
