Answer:
retupmoc
Explanation:
1.) Anwser will be retupmoc
because
public static String mysteryString(String s){
if(s.length() == 1){
return s;
}
else{
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
}
}
In this program input is "computer" . So the function mysteryString(String s) it does
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
so when it enters the first time ??s.substring(s.length() -1) and it will be give you 'r' then it calls the function recursively by reducing the string length by one . So next time it calls the mysteryString function with string "compute" and next time it calls return s.substring(s.length()-1)? + mysteryString(s.substring(0,s.length-1)) so this time it gives "e" and calls the function again recursively . It keeps on doing till it matched the base case.
so it returns "retupmoc".
Explanation:
==================
lin_spaced_vector.m
==================
function out=lin_spaced_vector(in1,in2)%defining function
out=linspace(in1,in2,200);%200 spaced numbers between in1 and in2
end
===================
Executable File
===================
clear all%clears history
clc%clears screen
lin_spaced_vector(1,10)%calling function
clear all
clc
lin_spaced_vector(1,10)
Answer:
Using unfamiliar abbreviations and acronyms
Explanation:
because he using RHFD and LTRE
Idunno if you got the answer to this question or if I'm too late, feel free to message me if I'm not!
Answer:
e. 4
Explanation:
The code segment will not work as intended if the value of the variable val is 4. This is because the while loop is comparing the value of the variable val to the value of each element within the numList. Since there is no element that is bigger than or equal to the number 4, this means that if the variable val is given the value of 4 it will become an infinite loop. This would cause the program to crash and not work as intended.