Create a class called Fraction. Provide a constructor that takes 2 integers. Provide methods for:
1 answer:
Answer:
Explanation;
else
System.out.println("f1 and f2 are not equal");
switch (input.charAt(0)
{
case '+':
f3 = f1.add(f2);
System.out.println("f1+f2=" + f3);
break;
case '-':
f3 = f1.subtract(f2);
System.out.println("f1-f2=" + f3);
break;
case '*':
f3 = f1.multiply(f2);
System.out.println("f1*f2="+f3);
break;
case '/':
f3 = f1.divide(f2);
System.out.println("f1/f2="+f3);
break;
default:
System.out.println("Illegal command: " + input );
break;
}
}// end of while loop
} // end of main
}
Note ; this is the last part of the programme check the attachment from 1-5 this is the 6th .
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Answer:
(1)Minterms complement = XYZ (2) Compliment of Minterms = Σm(0,1, 2 , 4 , 6) (3) (X+Y+Z) (4) Minimized SOP = Z + XY
Manterms = πM
Explanation:
Solution
Recall that:
Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)
(1) Canonical Disjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as minterms
In Minterm we assign 'I' to each uncomplimented variable and '0' to each complemented/complementary variable
For the given question stated we ave the following:
Minterms = XYZ, XYZ, XYZ, XYZ, XYZ.
(2) Canonical Conjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as maxterms.
In Maxterms we assign '0' to each uncomplimented variable and '1' to each complemented/complementary variable
Compliment of Minterms = Σm(0,1, 2 , 4 , 6)
Maxterms = πM
Note: Kindly find an attached copy of the complete solution to this question below.
