What range of addresses can be reached using the risc-v branch if equal (beq) instruction? (in other words, what is the set of possible values for the pc after the jump instruction executes? )

Define a class named movie. include private fields for the title,year, and name of the director. include three public functions withprototypes void movie: : settitle(cstring); , voidmovie: : setyear(int); , void movie: : setdirector(string); . includeanother function that displays all the information about a movie.write a main() function that declares a movie object namedmyfavoritemovie. set and display the object's fields.this is what i have but know its wrong since it will notcompile: #include#includeusing namespace std; //class declarationclass movie{private: string movietitle ; string movieyear; string directorname; public: void settitle(string title); void setyear(string year); void setdirector(string director); void displayinfo(); }; //class implementationvoid movie: : settitle(string title){ movietitle = title; cout< < "what is the title of themovie? "< > temp; myfavoritemovie.settitle(temp); cout< < "enter movie year"< > temp; myfavoritemovie.setyear(temp); cout< < "enter director'sname"< > temp; myfavoritemovie.setdirector(temp); //display all the data myfavoritemovie.displayinfo(); system("pause"); return 0; this code is not entirely mine someone on cramster edited my firstcode but then i try manipulating the new code and i still get acompile error message : \documents\visual studio 2008\projects\movie\movie\movie.cpp(46) : error c2679: binary '< < ' : no operator found which takes aright-hand operand of type 'std: : string' (or there is no acceptableconversion)c: \program files (x86)\microsoft visual studio9.0\vc\include\ostream(653): could be'std: : basic_ostream< _elem,_traits> & std: : operator< < > (std: : basic_ostream< _elem,_traits> & ,const char *)w

Computer programming is one type of what career

Use a software program or a graphing utility with matrix capabilities to solve the system of linear equations using an inverse matrix. x1 + 2x2 − x3 + 3x4 − x5 = 6 x1 − 3x2 + x3 + 2x4 − x5 = −6 2x1 + x2 + x3 − 3x4 + x5 = 3 x1 − x2 + 2x3 + x4 − x5 = −3 2x1 + x2 − x3 + 2x4 + x5 = 5