Answer:
The correct answer to the following question will be "Ribbon".
Explanation:
- In the configuration of the computer interface, the ribbon is indeed a graphical controlling feature in the format of a series of toolbars mounted on several tabs. The standard ribbon layout requires large toolbars, loaded with graphical keys, tabbed as well as other visual control features, organized by feature.
- The aim of the ribbon should be to provide quick and easy access to widely used activities for each system. The ribbon is therefore tailored for each task and includes program-specific commands. Besides, the top of a ribbon contains multiple tabs that can be used to disclose various groups of commandments.
Therefore, Ribbon it the suitable answer.
Answer:
"void" is the correct answer for the given question.
Explanation:
In the function body the value of balance variable is not return that means we use void return type .The void return type is used when the function does not return any value .
If the function are int return type that means it return "integer value ".
if the function are double return type that means it return the "double value" .
The complete implementation of this method is
public void deposit(double amount) // function definition
{
balance = balance + amount; // statement
}
Answer:
- Transform binary or unary M:N relationship or associative entity with its own key.
Explanation:
Transform binary relation is described as the method through which a decimal can easily be converted into binary while the unary relationship is described as a relationship in which both the two participants occurs from the same entity.
In the given case, 'transform binary or unary M:N relationship' can be created using 'the primary key linked with the relationship plus any non-key aspects of the relationship and the primary keys of the related entities' as it displays the existence of a relationship between the occurrences of a similar set of the entity i.e. associative entity here.
<span>void printArray(int a[ ]) {
</span> int i; <span>for (i = 0; i < a.length; i++){
</span> <span>System.out.print(a[i]);
</span> System.out.println(); <span>}
</span> }
Answer:
The class of language the machines recognise is Regular Language (See Explanation Below)
Explanation:
Given
Form δ : Q × Γ → Q × Γ × {R, S}
From the above transition form, it can be seen that the machine cannot read square symbols passed to it.
Regarding the square the machine is currently reading, there are multiple movement of S and it shouldn't be so because any number of the multiple movement can be simulated by exactly one movement of S.
As stated earlier that sequence of moves can be simulated by just one movement.
Let R = the movement
This means the machine can only use right move efficiently.
With this, we can say that the machine only read input string.
This is a characteristic of DFA (Deterministic Finite Automata).
With this, we can conclude that some DFAs will simulate the Turing machine and that they only read regular language.
