Assume that x is a variable that has been declared as a double and been given a value. Write an expression to compute the quartic root of x. The quartic root of a number is the square root of its square root. EXAMPLES: For example, the quartic root of 16.0 is 2.0 because: the square root of 16.0 is 4.0 and the square root of 4.0 is 2.0. Another example: the quartic root of 81.0 is 3.0 because the square root of 81.0 is 9.0 and the square root of 9.0 is 3.0. Thus, to find the quartic root of a number you take the square root of the number and then take the square root of that. In this exercise you must find the quartic root of x in a single expression-- you must not write any statements. Also, you may only use the sqrt() function-- no other functions. (HINT: you will need to call the sqrt() function twice-- and you will need to pass the return value of one of those calls as argument to the other call. AND REMEMBER: write an expression, not a statement.)

By -
.
.
Click on the title for the solution
.
.

This is the answer:

:



sqrt(sqrt(x))

Comments

Post a Comment

Popular posts from this blog

411: You are given a file named phonedir that consists of many lines containing three strings: lastname firstname emailaddress. Write a utility program that reads commands (from stdin). Each command has one of two possible forms: lookup string add lastname firstname emailaddress In the case of the first command (lookup), the program looks for a line in the phonedir file where either the firstname or lastname matches the string given. If it finds such a line it prints out all three parts of the line, separated by spaces: lastname firstname emailaddress If it does not find a match in the file it prints out the string it was looking for, followed by a colon followed by a space followed by the message "not found": string: not found In the case of the second command (add), the program appends the three strings given to the file phonedir. Hints and suggestions: (1) Define and use two functions named lookup and add. When your program reads the string lookup, your lookup function is called; when the program reads the string add, your add function is called. When either of these two functions are called, they then read whatever else is necessary for their command (one string for lookup and three strings -- lastname, firstname, and emailaddress-- for add). And then these functions do... whatever it takes to carry out the command. (2) When doing a lookup or an add, open up the file phonedir ... carry out the operation .. and then close up the file. EXAMPLE: suppose the phonedir file looked like this: arnow david arnow@panix.com bush george president@whitehouse.gov gates bill bill@microsoft.com here then is a sample session with the program (program output is in bold): lookup david arnow david arnow@panix.com lookup joe joe: not found add theplumber joe joetheplumber@nowhere.com lookup arnow arnow david arnow@panix.com lookup joe theplumber joe joetheplumber@nowhere.com and the phonedir file would now be: arnow david arnow@panix.com bush george president@whitehouse.gov gates bill bill@microsoft.com theplumber joe joetheplumber@nowhere.com

By Author -
. . Click on the title for the solution . . This is the answer: : #include <iostream> #include <fstream> #include <string> using namespace std; void lookup(); void add(); int main() { string type; while (cin >> type) { if (type == "lookup") lookup(); else  add(); } return 0; } void lookup() { ifstream fin; fin.open("phonedir"); if(!fin.is_open()) cout << "Not Open" << endl; bool check = false; string name,firstName,lastName,email; cin >> name; while (fin >> firstName >> lastName >> email) { if (firstName== name || lastName==name) { cout<<firstName<<" "<<lastName<<" "<<email<<endl; check=true; } } if(!check) cout << name << ": not found" << endl; fin.close(); } void add() { ofstream fout; string name; string fir...
Read more

NOTE: in mathematics, division by zero is undefined. So, in C++, division by zero is always an error. Given a int variable named callsReceived and another int variable named operatorsOnCall write the necessary code to read values into callsReceived and operatorsOnCall and print out the number of calls received per operator (integer division with truncation will do). HOWEVER: if any value read in is not valid input, just print the message "INVALID".

By -
. . Click on the title for the solution . . This is the answer: : cin>>callsReceived>>operatorsOnCall; if (operatorsOnCall>0) {cout<<callsReceived/operatorsOnCall;} else cout<<"INVALID";
Read more