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3 years ago

Which document lists the planned dates for performing tasks and meeting goals identified in the project plan?

1 answer:

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<span>The document is called as Project schedule. this document contains the tasks that needs to be performed, the time that is given to the perform the task, the target dates. It also contains the budget of the project and the budget can be represented task wise. This helps in performing the project in a more organized way.</span>

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a third important logical element is the inverter. an inverter does pretty much what it says. if the input is 0, the output is 1

Mrrafil [7]

Answer:

$xy225.252 < \geqslant > . \times 5 \times . = {?}^{2}$

7 0

3 years ago

If you want to adjust the width of a column manually, move your mouse over the _______ column boundary and click and drag until

Luda [366]

Answer:

The answer is "Right".

Explanation:

In computer science, editing is the primary activity, which is used by the users to handles graphics, tables, and other things. It can be divided into two types first is internal(default) and the second is manual.

In the manual editing, to edit the column of the table we just right click on the column.

It enables you to edit option, in which we edit column according to our desire.

6 0

3 years ago

What are at least three tips to taking photographs of insects? Why would following the tips help create better photographs?

Ber [7]

1) Use a good cam 2) Make sure you focus on it 3) Make sure you snap it at the right time.... I hope this helps!!!!!

7 0

3 years ago

Can run mobile-style apps in the area known as the start screen, while running traditional software in the more familiar desktop

Travka [436]

The correct answer: Yes, mobile-style apps can run in a personal computer's desktop.

That is possible by means of a desktop application called emulatator. An emulator like Bluestacks allows a personal computer to run mobile-style apps by acting as a virtual drive in the personal computer's harddisk.

Emulation is successful if the system requirements of the mobile-application is met by the personal computer's system attributes such as Random Access Memory abundance, Random Access Memory speed, Processing speed (in some cases core abundance e.g. core 2) etc.

Some mobile-applications do not work in the personal computer's desktop, however, if this application requires platform specific functions such as mobile device's network provider etc.

4 0

4 years ago

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.

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3 years ago