Answer: Logical Point Blocking
Explanation:
Answer:
False.
Explanation:
A stethoscope is a medical instrument which is used for listening to the work of the heart and lungs. It consists of a microphone that rests on the patient's chest, and rubber tubes that bring the sound to the earphones. An ordinary stethoscope does not have any moving or electrical parts, as it only mechanically conducts sound from the patient's chest to the doctor's ear. One of the simplest examinations that doctors perform today is listening to the lungs and heart with a stethoscope. With the help of this instrument, noises are heard during the work of the lungs and heart.
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: C because bill gates is the owner
Answer:
a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1
Explanation:
a. How many bits are needed for the opcode?
Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72
72 = 64 + 8 = 2⁶ + 8
Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128
So, we require 7 bits for the opcode.
b. How many bits are left for the address part of the instruction?
bits left = bits per word - opcode bit = 16 - 7 = 9 bits
c. What is the maximum allowable size for memory?
Since there are going to be 2⁹ bits to addresses each word and 16 bits for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.
We convert this to bytes
8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB
d. What is the largest unsigned binary number that can be accommodated in one word of memory?
Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus
2¹⁶ - 1
