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ioda
3 years ago
6

Which of the following is the output of the code

Computers and Technology
1 answer:
olya-2409 [2.1K]3 years ago
3 0

Answer:

need to see code

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Decrease in error CAD software makes use of some of the best tools, the percentage of error that occurred because of manual designing is significantly reduced.

Decrease in effort when it comes to the amount of effort that was needed for the sake of designing the different models, it has been reduced significantly because the software automates most of the task.

Saves time when you are using the computer aided design software, it will save your time and you can make better and more efficient designs in shorter time duration.

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3 years ago
What is a computer?write any four features of computer​
igomit [66]

Answer:

A computer is defined as an electronic device for storing and processing data, typically in binary form, according to instructions given to it in a variable program.

Four features of a computer would be CPU, GPU, Memory, and Motherboard.

8 0
2 years ago
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Hey, Another question. I'm sure it's possible in the future, but I wanted to ask if HIE would be possible. I'm sure it would be,
dolphi86 [110]

Answer:

Originally Answered: Will there ever be a band as big as the Beatles? No, there will never be a band as popular as the Beatles. For one thing they were exceptionally good in a time of great music in general.

Explanation:

please mark this answer as brainliest

7 0
3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
3 years ago
Is it possible to build a portal in real life?
san4es73 [151]

No because we are not that  far in technology.

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3 years ago
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