The answers are as follows:
a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC
= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)
= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC
= B'(C' + C) + B(C' + C) = B' + B = 1
b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms
with x'1, which can be factored and removed as in (a). The remaining 2n1
product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms
with x'2, which and be factored to remove x2 and x'2, continue this
process until the last term is left and xn + x'n = 1
The code below is written in javascript which gives us the integer,
<h3>The function code is</h3>
function solution (num){
//Javascript function
var x=0,sum=0,a,b;
var d=""; //variable declaration
var digits = num.toString().split('');
var Individual = digits.map(Number);
//console.log(Individual);
for (var i=0;i<Individual.length;i++)
{//For loop
x=x+Individual[i];
}
var y=x*2;
for(var i=0;i<Individual.length;i++)
{
sum=sum+Individual[i];
d=""+d+Individual[i];
if(sum==y)
break;
if(i==Individual.length-1)
{
i=-1;
continue;
}
}
console.log("the value is "+d);
}
var number = prompt("enter the number");
//Asking user for value...
console.log("the number is");
console.log(number);
solution(num
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It's often Linux since Linux is free, but it could be Windows.
<span>The answer is False. When declaring an array, we use 1 set of square brackets.</span>
