Answer:
For values when n<32 use fb(n) else use fa(n).
See explaination for details
Explanation:
Earlier when n=1 fa(n)=2000 while fb(n)=2. for n=2 fa(n)=2000*22 =2000*4=8000 while fb(n)=2*24=2*16=32. It is observed that fa(n) requires more time than fb(n) for small values.
Now, we will see when fb(n) crosses fa(n). This can happen only when fb(n) values equals or greater than fa(n)
therefore,
2000n2<=2n4
Solving equation we get n2>=1000 which can happen when n>=32.
So for values when n<32 use fb(n) else use fa(n)
Answer:
200 Ω
Explanation:
Hi there!
Please see below for the circuit diagram.
<u>1) Find the total resistance of the resistors in parallel</u>
Total resistance in parallel equation: 
Both the resistors measure 200 Ω. Plug these into the equation as R₁ and R₂:
Therefore, the total resistance of the resistors in parallel is 100 Ω.
<u>2) Find the total resistance of the circuit</u>
Now, to find the total resistance of the circuit, we must add the 100 Ω we just solved for and the 100 Ω for the other resistor placed in series:
100 Ω + 100 Ω = 200 Ω
Therefore, the total resistance of the circuit is 200 Ω.
I hope this helps!
Answer:
0
Explanation:
As far as I know, no one I know, even knows that brainly exists. Those who do know about it don't use it.
Answer:
public class SimpleSquare{
public int num;
private int square;
public SimpleSquare(int number){
num = number;
square = number * number;
}
public int getSquare(){
return square;
}
}
Explanation:
*The code is in Java.
Create a class called SimpleSquare
Declare two fields, num and square
Create a constructor that takes an integer number as a parameter, sets the num and sets the square as number * number.
Since the square is a private field, I also added the getSquare() method which returns the value of the square.
