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tamaranim1 [39]
3 years ago
10

HELP ASAP Encoding a video format and then decoding it during playback is one of the functions of MPEG-4 and H.264 file players.

MPEG-4 and H.264 are known as. (A) video editors. (B) video codecs. (C) video streamers. (D) video monitors. GIVING 7 or 14 points for brainliest
Computers and Technology
2 answers:
Mademuasel [1]3 years ago
5 0

Answer: video codects

Explanation:

Nataly_w [17]3 years ago
5 0

Answer:

B

Explanation:

edg21

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8.3 code practice edhesive PLEASE HELP AND HURRY
NemiM [27]

Answer:

numbers = '14 36 31 -2 11 -6'

nums = numbers.split(' ')

for i in range(0, len(nums)):

  nums[i] = int(nums[i])

print(nums)

8 0
3 years ago
Does anyone know how many Brainliests you need to be able to send a private message to someone??
White raven [17]

Answer:

no idea sorry...

Explanation:

8 0
3 years ago
Read 2 more answers
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
)What item is at the front of the list after these statements are executed?
Dominik [7]

Answer:

C.Rudy

Explanation:

A deque waitingLine is created.Then Jack is added to front of the deque first.After that Rudy is added to the front of the deque.AFter that Larry and sam are inserted at the back.So the last item inserted at the front is front of the deque.So Rudy was the last item inserted at the front.

Hence we conclude that the answer is Rudy.

4 0
3 years ago
Why are tariffs and other trade barriers economically harmful when they save some jobs
Dmitry [639]
Tariffs economically harmful save they save some jobs because higher prices are forced upon the people by the tariffs will cost more jobs than they save.

7 0
3 years ago
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