Question

C++ help

Answer 1

#include <iostream>;
#include <iomanip>;
#include <string>
using namespace std;

int main()
{
int yearRate, payments;
double loan, monthPay, paidBack, interest, rate;

cout << setprecision(2) << fixed;
//gets loan amount,
cout << "Enter the loan amount => ";
cin >> loan;
//gets yearly interest rate
cout <<"Enter the yearly interest rate as a percentage => ";
cin >> yearRate;
//gets number of payments
cout << "Enter the number of payments ";
cin >> payments;

//This is where the formula goes

//outputs the loan, yearRate, payments, monthPay, paidBack, and interest variables
cout << "Loan Amount: $ " << loan;
cout << "Yearly Interest Rate: $ " << yearRate;
cout << "Number of Payments: $ " << payments;
cout << "Monthly Payment: $ " << monthPay;
cout << "Amount Paid Back: $ " << paidBack;
cout << "Interest Rate: $ " << interest;

//option to end program
cin.ignore();
cout << "Press enter to end";
cin.get();
return 0;
}

Answer 2

Payment = (Rate * (1 + Rate)^N)/(((Rate + 1)^N) - 1)

Answer 3

Payment is the monthPay variable. Rate is the rate variable. N is the payment variable. L is the loan variable.

Answer 4

I need to turn the "//This is where the formula goes" section into a c++ formula that is the same as the formula i listed

Answer 5

ther is no ^ operator. There is a pow function in cmath

Answer 6

i know, the formula i listed is not cmath, i need to convert it

Answer 7

the only way that i can incorporate the exponents is by making another variable

Answer 8

from what i can tell

Answer 9

what's wrong with rate * pow((1+rate),n) etc.

Answer 10

\[Payment = \frac{Rate * (1 + Rate)^{N} }{ ((1 + Rate)^{N} - 1} * L\]
this is how the original formula looks

Answer 11

umm i don't know, does that work?

Answer 12

btw there is a ) after the -1 i forgot to add

Answer 13

thx for ur help, that does seem to work

Answer 14

\[Payment = \frac{Rate * pow((1 + Rate),N) }{ pow((1 + Rate),N) - 1} * L\]

yep. no problem