Here's a solution that works except for the leading zeros. Unclear (to me) why they need to be there and what's the logic?
public static void main(String[] args)
{
int number, base;
Scanner Cin = new Scanner(System.in);
System.out.print("Enter a decimal number between 1 and 100,000: ");
number = Cin.nextInt();
System.out.print("Enter a base from 2-16: ");
base = Cin.nextInt();
System.out.format("The number %d in base %d is %s.", number, base, Integer.toString(number, base).toUpperCase());
}
Answer:
<h3>A. Is used to add a personal touch to e-mail messages.</h3>
Explanation:
<h3>A digital signature—a type of electronic signature—is a mathematical algorithm routinely used to validate the authenticity and integrity of a message (e.g., an email, a credit card transaction, or a digital document). ... Digital signatures are significantly more secure than other forms of electronic signatures.</h3>
Answer:
(A) Always true
<em></em>
Explanation:
Given
Boolean variables x and y
Required
What is (x && y) || !(x && y)
Irrespective of what x or y is, the x && y is always true
Take for instance:
<em>x = true;</em>
<em>y = true;</em>
<em>x&&y will also be true because the values of x and y were not altered</em>
<em />
<em>And this is true for whatever boolean value x or y assume</em>
<em />
Having said that:
x&&y = true
So, the expression is analysed as follows:
(x && y) || !(x && y)
Substitute true for x&&y
(true) || !(true)
<em>!(true) = false</em>
So, the expression becomes:
true || false
|| represents the OR operator; and it returns true if at least one of the conditions is true.
By this:
true || false = true
<em>Option (A) Always true </em><em>answers the question.</em>
Answer:
Proofreading means carefully checking for errors in a text before it is published or shared. It is the very last stage of revising a text, when you fix minor spelling and punctuation mistakes, typos, formatting issues and inconsistencies.
