Answer:
The answer to this question defined below.
Explanation:
It's a smart idea to get a common language for coding of every kind. It would help all developers and customers understand the language better because, in every case, there's no more need to learn, that language.
- This could also render software developed in the very same language consistent, and therefore, ports on multiple platforms are not required.
- In this process, we talk about the common property and function of the classes, that's why it is the correct answer.
In the crankcase
I hope this helps
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
the answer is going to be c
Answer: 83.17
Explanation:
By definition, the dB is an adimensional unit, used to simplify calculations when numbers are either too big or too small, specially in telecommunications.
It applies specifically to power, and it is defined as follows:
P (dB) = 10 log₁₀ P₁ / P₂
Usually P₂ is a reference, for instance, if P₂ = 1 mW, dB is called as dBm (dB referred to 1 mW), but it is always adimensional.
In our question, we know that we have a numerical ratio, that is expressed in dB as 19.2 dB.
Applying the dB definition, we can write the following:
10 log₁₀ X = 19.2 ⇒ log₁₀ X = 19.2 / 10 = 1.92
Solving the logarithmic equation, we can compute X as follows:
X = 10^1.92 = 83.17
X = 83.17
