Answer:
A. f/2.8
Explanation:
A hole within a lens, through which light travels into the camera body is referred to as the APERTURE.
It is typically expressed as "f number" in photography. Basically, a fast lens is any lens with a maximum aperture of f/4 or more i.e f/2.8 , f/1.8 , f/1.4 etc.
The smaller the number is the bigger the maximum aperture is. Hence, the bigger the maximum aperture the more light that your lens will allow in.
Therefore, the ideal aperture settings for a fast lens is f/2.8
The answer is Object-oriented programming (OOP). This is a type of programming language model on the concept of "objects", which may comprise of data, in the form of fields, often known as attributes; and code, in the form of procedures, often known as methods. The first step here is to classify all the objects the programmer wants to work and how they relate to each other, this is known as data modeling. Once an object has been identified, it is generalized as a class of objects which describes the type of data it comprises and any logic sequences that can manipulate it.
Answer:
Category 3 (CAT3) Unshielded Twisted Pair cable
Explanation:
A Category 3 (CAT3) Unshielded Twisted Pair cable is a cable that makes use of the conductivity of copper to transmit data and power. These cables are an older standard for Ethernet cables and can handle data speeds of up to 10mbps.The succeeding standard for Ethernet cables include CAT5 which has speeds of up to 100 mbps, CAT 3 is mostly used in applications where speed is not of great importance such as in PBXs and VoIP telephone systems.
Answer:
It can be 100001, 100000,000001,000000
Hence, answer is 4.
its also can be found through 2C2 *2 = 2!/0! *2 =2*2 =4
Explanation:
The answer is straight forward, we have 2 empty places and we can select from 2 items, and hence its 2C2, and now we can arrange them as well, and since we can arrange in 2 ways, hence a total number of possibility = 2 * 2C2 = 4, which is the required answer.
Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
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Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
