The design process is the same in all creative occupations, thus, similar skills and qualities are required. The work of an architect, Craftperson, Fine Artist, or an Interior designer is to design. Whether one occupation designs clothing and associated fashions or another one designs interiors of building, they all design. Most people who work as designers possess similar personal qualities like creativity, better communication skills, and high skill set.
Answer:
Option B is the correct answer.
Explanation:
- In the above code, the loop will execute only one time because the loop condition is false and it is the Do-While loop and the property of the Do-while loop is to execute on a single time if the loop condition is false.
- Then the statement "x*=20;" will execute one and gives the result 200 for x variable because this statement means "x=x*20".
- SO the 200 is the answer for the X variable which is described above and it is stated from option B. Hence it is the correct option while the other is not because--
- Option A states that the value is 10 but the value is 200.
- Option C states that this is an infinite loop but the loop is executed one time.
- Option D states that the loop will not be executed but the loop is executed one time
Answer:
#include <iostream>
using namespace std;
double DrivingCost(int drivenMiles,double milesPerGallon,double dollarsPerGallon)
{
double dollarsperMile=dollarsPerGallon/milesPerGallon;//calculating dollarsperMile.
return dollarsperMile*drivenMiles;//returning thr driving cost..
}
int main() {
double ans;
int miles;
cout<<"Enter miles"<<endl;
cin>>miles;
ans=DrivingCost(miles,20.0,3.1599);
cout<<ans<<endl;
return 0;
}
Output:-
Enter miles
10
1.57995
Enter miles
50
7.89975
Enter miles
100
15.7995
Explanation:
In the function first I have calculated the dollars per mile and after that I have returned the product of dollarspermile and driven miles.This will give the cost of the Driving.
Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
