Need with beginner level c++. "a user types a word and a number on a single line. read them into the provided variables. then print: word_number. end with newline. example - if user entered amy 5, the output should be: amy_5 ." anyone able to explain how to do this? #include #include using namespace std; int main() { string userword; int usernum; // put your solution here. return 0; }

The camera still is bad even with the new iphone xr and especially in low light it is even worst because you can see the pixels more if its in low light. if all you apple customers want apple to fix this then lets fill there feedback with complaints about the

Use a physical stopwatch to record the length of time it takes to run the program. calculate the difference obtained by calls to the method system. currenttimemillis() just before the start of the algorithm and just after the end of the algorithm. calculate the difference obtained by calls to the method system. currenttimemillis() at the start of the program and at the end of the program so that the elapsed time includes the display of the result. use the value returned by the method system. currenttimemillis() just after the end of the algorithm as the elapsed time.

Write a new lc-3 trap subroutine (i. e. a subroutine that will be invoked via the trap instruction) that will receive a numeric digit entered at the keyboard (i. e. an ascii character), echo it to the screen, and return in r0 the corresponding numeric value: so if the user types the digit '7', the character '7' will appear on the screen, but the value returned in r0 will be b0000 0000 0000 0111 (#7) you may not use any trap calls in your code - you must implement the "polling" code that interrogates the keyboard status and data registers. ; getnum_tsr ; a subroutine for obtaining a numeric value ; given ascii numeric digit input to keyboard. ; the numeric digit is echoed to the console (e. g. '7' = b0000 0000 0011 0111), ; but the value returned in r0 is the actual numeric value ; corresponding to the digit (e. g. b0000 0000 0000 0111 =