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Approximately 84,800,000 barrels per day.
The best produce more than 100,000 barrels per day (1 barrel=42 gallons). Most produce less than this, and can produce as little as 100 barrels per day. A "good" well comes down to economics. It should be making enough money to offset the time, effort, and energy costs to keep it running and be producing sufficently to warrent keeping the equipment dedicated to it. Otherwise the equipment should be moved to a more economically viable well which will yield a greater profit margin.
#define leap year void main() { enumeratedmonth{jan=1, feb, mar, april, may, june, july, aug, sept, oct, nov, dec} enumeratesd month month; int day,days,a month clrscr() printf("enter no.of month"); scanf("%d",& a month); switch(a month) { case jan: case mar: case may: case aug: case oct: case dec: printf(" no. of day=31"0; days=31; break; case april: case june: case sept: case nov: days=30; break; default; printf("wrong no."); day=0; } if (day!=0) printf("no. of days=%d/n",days); if (a month==feb) printf("29 if its a leap year"); getch(); }
The following is one possible method. If the input arguments would result in an invalid date (or no arguments are given), then the members are set to an arbitrary default date (in this case 1/1/1980). An alternative method that ensures all dates are valid (non-default) would be to throw an exception from the set() member. If your code is producing invalid date arguments (causing the exception to be thrown) then you really need to fix the code that generates those dates. Note that the static methods, is_valid() and is_leap_year(), can be used to check date arguments without actually constructing a date object. Another alternative would be to set a flag that denotes if the date object is valid or not. However, this is the least intuitive method as the onus is then placed upon the user to always check the flag before using the object. #include<iostream> #include<iomanip> class cdate { unsigned m_day; unsigned m_month; unsigned m_year; public: cdate (unsigned day, unsigned month, unsigned year): m_day(1), m_month(1), m_year(1980) { set (day, month, year); } cdate (const cdate& date): m_day(date.m_day), m_month(date.m_month), m_year(date.m_year) {} cdate& operator= (const cdate& date) { m_day=date.m_day; m_month=date.m_month; m_year=date.m_year; } bool set (unsigned day, unsigned month, unsigned year); static bool is_valid (unsigned day, unsigned month, unsigned year); static bool is_leap_year (unsigned year); unsigned day() const { return m_day; } unsigned month() const { return m_month; } unsigned year() const { return m_year; } }; bool cdate::is_valid (unsigned day, unsigned month, unsigned year) { // handle zeroes if (!day !month !year) return false; // handle invalid month if (month>12) return false; // handle 31 day months if ((month==1 month==3 month==5 month==7 month==8 month==10 month==12) && day>31) return false; // handle 30 day months if ((month==4 month==6 month==9 month==11) && day>30) return false; // handle leap years if (month==2) { if (!is_leap_year (year) && day>28) return false; else if (day>29) return false; } // data ok return true; } bool cdate::is_leap_year(unsigned year) { // common year (not divisible by 4) if (year%4) return false; // leap year (not divisible by 100) else if (year%100) return true; // leap year (divisible by 400) else if (year%400==0) return true; // common year return false; } bool cdate::set (unsigned day, unsigned month, unsigned year) { if (is_valid (day, month, year)) { m_day = day; m_month = month; m_year = year; return true; } // alternatively, throw an exception here return false; } std::ostream& operator<< (std::ostream& os, const cdate& date) { os << std::setw (2) << std::setfill('0') << date.day() << '/' << std::setw (2) << std::setfill('0') << date.month() << '/' << std::setw (4) << std::setfill('0') << date.year(); return os; } int main () { // invalid date (was not a leap year) std::cout << "Input:\t29/02/1900" << std::endl; cdate d1(29,2,1900); std::cout << "Output:\t" << d1 << std::endl; // valid date (was a leap year) std::cout << "Input:\t29/02/2000" << std::endl; cdate d2(29,2,2000); std::cout << "Output:\t" << d2 << std::endl; }
Hi gys, are you looking to write a program to know the sign depends on the birth date or horoscope on python? write down following codes on python, those will make a nice program. #This program will get the horoscope sign. def getSign(): name = raw_input ("Enter Your Name: ") month = raw_input("Enter Birth Month: ") day = int(raw_input("Enter Birth Day of Month: ")) if month "y": main() letter = raw_input("Go again? (y/n): ")
For the US, the average is somewhere between 15 barrels a day and 30 barrels a day, but some are producing much more.
they produce about 400,000
depending on how much they work in a month they could make roughly $33 000 a month, you would get about that much if you work 1 hour a day
Every month that is not February has a 28th day in it. Those months also have a 29th day, a 30th day, and some have a 31st day as well.
it produce about 456 time a day
a gallon
a bathtub!
65poo
approx. 2 liters a day.
5lbs per person per day
About a gallon a day .
National Mail day, National underpants day, National Stupid Question day. April Fools! All of those were fake.