To make a horizontal line in mathematics you require some sort of constant function. Where no matter the input x the output y will always equal to a certain number n.
For eg. consider n = 2. Hence the graph of line has an equation of y = n making it always intercept in (x, n). As x increases the y is the same therefore the horizontal infinite line emerges.
So to plot or code this kind of program you need to have the same values of y therefore y is a constant function.
Hope this helps.
r3t40
Import java.util.Scanner;
class hola
{
public static void main(String[]args)
{
Scanner x=new Scanner(System.in);
int a=x.nextInt();
int b;
if(a>20&&a<100)
{
b=a%12;
if(b%2==0){
System.out.print("es par"+b);
}
else{
System.out.print("es impar"+b);
}
}
}
}
Answer:
a) Yes
b) Yes
c) Yes
d) No
e) Yes
f) No
Explanation:
a) All single-bit errors are caught by Cyclic Redundancy Check (CRC) and it produces 100 % of error detection.
b) All double-bit errors for any reasonably long message are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
c) 5 isolated bit errors are not caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit since CRC may not be able to catch all even numbers of isolated bit errors so it is not even.
It produces nearly 100 % of error detection.
d) All even numbers of isolated bit errors may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
e) All burst errors with burst lengths less than or equal to 32 are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
f) A burst error with burst length greater than 32 may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit.
Cyclic Redundancy Check (CRC) does not detect the length of error burst which is greater than or equal to r bits.
Answer:
press flash
Explanation:
its on the bottom row on the remote on mine
Hi, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.
Answer:
import math
def rectangle(perimeter, area):
l1_1 = (perimeter+math.sqrt((perimeter**2)-(16*area)))/4
l1_2 = (perimeter-math.sqrt((perimeter**2)-(16*area)))/4
l2_1 = area/l1_1
l2_2 = area/l1_2
print(l1_1,l2_1)
print(l1_2,l2_2)
if l1_1.is_integer() and l2_1.is_integer() and l1_1>0 and l2_1>0:
return(int(max(l1_1,l2_1)))
elif l1_2.is_integer() and l2_2.is_integer() and l1_2>0 and l2_2>0:
return(int(max(l1_2,l2_2)))
else:
return(None)
Explanation:
- We import math to make basic operations
- We define the rectangle function that receives perimeter and area
- We calculate one of the sides (l1_1) of the rectangle using the quadratic equation to solve 2h^2 - ph + 2a = 0
- We calculate the second root of the quadratic equation for the same side (l1_2)
- We calculate the second side of the rectangle using the first root on w = a/h
- We calculate the second side of the rectangle using the second root on w= a/h
- We verify that each component of the first result (l1_1, l2_1) is an integer (using the build-in method .is_integer) and greater than 0, if True we return the maximum value between them (using the max function) as w
- If the first pair of sides evaluate to False we check the second root of the equation and if they meet the specification we return the max value
- if all the if statements evaluate to false we return None to indicate that not positive or integer sides were found
