Using a financial statements model like the one shown below, record the appropriate amounts for the following two events: (In the Cash Flow column, use the initials OA to designate operating activity, IA for investing activity, FA for financing activity, and NC for net change in cash. Amounts to be deducted should be indicated with minus sign. Leave no cells blank - be certain to enter \"0\" wherever required. Omit the \"$\" sign in your response.) January 1, 2010, issue of the note payable. December 31, 2010, payment on the note payable. Solution (1) assets( +cash ) = liability ( + notes payable) = FA statement of cash flow (2) assets(- cash) = liability (- notes payable) = FA \" = $0 Net cash.

1. Using a loop structure, write function myPow (double x, int y) which calculates x^y. Write a
C++ program that uses myPow() to calculate the following: z = x^4 + 2x^3 + 6x - 2 + 2/x^2 +
3/x^3 where x is entered by the user Sample output: Enter x: 2 z = 42.875 Press any key to
continue ... Modify your program such that the myPow(double x, int y) function can handle the
case when y is negative. Test your program using the following equation: k = x^-3 + 2x^-2
Sample output: Enter x: 2 z = 42.875 k = 0.625
Solution
#include
#include
using namespace std;
double myPow(double x, int y)
{
// base case
if (y == 0)
return 1;
// base case
else if (y == 1)
return x;
int power;
if(y > 0)
power = y;
else
power = -y;
int copypower = power;
// positive x
double input;
if(x > 0)
input = x;
else
input = -x;
// intially answerr is x
double answer = input;

2. int temp = 1;
while (power / 2 > 0)
{
answer = answer * answer;
power = power / 2;
temp = temp * 2;
}
// recursively calling the function
answer = answer * myPow(input, copypower - temp);
// to handle negative answer
if (x < 0 && y % 2 == 1)
answer = -answer;
// to handle negative power
if (y < 0)
answer = 1 / answer;
return answer;
}
// Driver program to test above functions
int main()
{
double x;
printf("Enter x: ");
scanf("%lf",&x);
double z = myPow(x,4) + 2*myPow(x,3) + 6*x - 2 + (2/myPow(x,2)) + (2/myPow(x,3));
double k = myPow(x,-3) + 2*myPow(x,-2);
printf("z: %0.3lf ",z);
printf("k: %0.3lf ",k);
return 0;
}

3. /*
output:
Enter x: 2
z: 42.750
k: 0.625
*/