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2 years ago

5

Harry needs to make a presentation on endangered species of animals. Which presentation software elements can he use to display

animals along with their reason of extinction?
A. image and flowchart
B. image and table
C. text and table
D. shapes and image
E. graph and shapes

Computers and Technology

1 answer:

antiseptic1488 [7]2 years ago

4 0

I Think The answer is either b or d

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Write the definitions for three function named max. Each receives two parameters, of the same type, and returns the larger of th

GenaCL600 [577]

Answer:

The method definition to this question can be given as:

Method definition:

double max(double x, double y) //define method with double parameter

{

if (x>=y) //check condition.

return x; //return value

else

return y; //return value

}

double max(int x, int y) //define method with integer parameter

{

if (x>=y) //check condition

return x; //return value

else

return y; //return value

}

double max(char x, char y) //define method with char parameter

{

if (x>=y) //check condition

return x; //return value

else

return y; //return value

}

Explanation:

The above method definition can be described as below:

In the first method definition first, we define a method that is "max()". In this method we pass two variables as a parameter that is "x and y" and the datatype of this is double. Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
In the second method definition, we define a method that is same as the first method name but in this method, we pass two integer variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
In the third method definition, we define a method that is same as the first and second method name but in this method, we pass two char variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.

8 0

3 years ago

Which statement outputs a double value in a field of nine characters with three digits after the decimal point?

zimovet [89]

Answer:

Console.WriteLine("Format Double: {0:n3}", num); //formatting output with 3 digit decimal point

Explanation:

Following are the program in c#

using System; // namespace

using System.Collections.Generic;

using System.Linq;

using System.Text;

namespace Tasks

{

class Program 2 // program2

{

static void Main(string[] args) // Main function

{

double num = 958254.73789621; // variables

Console.WriteLine("Format Double: {0:n3}", num); //formatting output with 3 digit decimal point

Console.Read();

}

}

}

Output:

Format Double : 958254.737

Here we have declared a variable num of type double which store the value num=958254.73789621. To do format with the double number i used a syntax {0:n3}. This syntax {0:n3}is separated with :(colon) here 0 represent the value before the decimal point that is 958254 and n3 represent the value upto 3 decimal points. Hence this statement give the output with three digit after the decimal point .

3 0

3 years ago

What are some things that games were historically used for?

madreJ [45]

Answer:

D) All of the above

Explanation:

All of these options are true.

Hope it helps and is correct!

5 0

3 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.

3 0

2 years ago

2. The part of the computer that provides access to the Internet is the A. monitor. B. keyboard. C. system unit. D. modem.

allochka39001 [22]

The modem provides access to the internet.

4 0

3 years ago

Read 2 more answers