Answer:
1. G=D+(A+C^2)*E/(D+B)^3
cobegin:
p1: (D+B)
p2: p1^3
p3: C^2
p4: A+ p3
p5: E/p2
p6: p4 * p5
p7: D + p6
:G
coend
2. Now The value A=2, B=4, C=5, D=6, and E=8
p1: 6+4 =10
p2: p1 ^3= 10^3= 1000
p3: c^2= 5^2 =25
p4: A + p3= 2 +25 =27
p5: 8/1000
p6: 27 *8/1000
p7: D+ P6= 6+ 216/1000
= 6216/1000
=6.216
Explanation:
The above, first bracket with power is processed, and then power inside and outside bracket. And rest is according to BODMAS, and one process is solved at a time.
Answer:
Programming languages to help solve algorithms
Explanation:
Answer:
The first choice
Explanation:
I took the test and that was the correct answer
Hope it helped!
Answer:
// This program is written in C++ programming language
// Comments are used for explanatory purpose
// Program starts here
#include <iostream>
#include <string>
using namespace std;
// Declare variables
int inputvar;
// Declare output variable as array
int outputvar[32];
// Set a counter for binary array
int i = 0;
while (inputvar > 0) {
// Divide inputvar by 2 and store remainder in outputvar
outputvar[i] = n % 2;
inputvar/=2;
i++; // increment i by 1
}
// End of division
// Prin resulting array in reverse order
for (int j = i - 1; j >= 0; j--) {
cout << outputvar[j];
}
return 0;
}
// End of Program
Answer:
C. sqrt(Math)
Explanation:
All but one of options A to E are is not a static method.
Only option C is a static method. The sqrt() is a static method of Math, that can always be used as Math.sqrt() is used;
The Math class defines all of its methods to be static. Invoking Math methods is done by using Math as a method rather than a variable of type Math; this means that sqrt(Math) doesn't rely on instance variables and don't need to be overridden, unlike others.
Lastly, sqrt(Math) is a static method because unlike other options, it is an utility method, and it is relevant to computations on primitive data types.
The purpose of the static method is in large part to offer a standard library of functions, and it doesn't need to be applied directly to an object.
