<span>It is a logical, systematic search for the source of a problem in order to solve it, and make the product or process operational again.Troubleshooting is needed to identify the symptoms. ...Troubleshooting is the process of isolating the specific cause or causes of the symptom.</span>
The answer is c
just replace the y with the y value and the x with the x value
Calculate LER for a rectangular wing with a span of 0.225m and a chord of 0.045m. The weight of the glider is 0.0500 Newtons. (Note: the wing span is the width of the wing and is measured from wing tip to wing tip, or perpendicular to the fuselage. The wing chord is the length of the wing measured parallel or along the length of the fuselage.)
Answer:
Area of rectangular wing = span × chord = 0.225×0.045= 0.010125m2
LER = Area/weight = 0.010125/0.0500 = 0.2025
Answer:
//Code is created using java
import java.util.*;
// returns the sum
public int sum(int N)
{
if(N==1)
return (1);
else
return N+sum(N-1);
}
// code to return the Bipower ouput
public int BiPower(int N)
{
if(N==1)
return (2);
else
return 2*BiPower(N-1);
}
// Code to return TimesFive output
public int TimesFive(int N)
{
if(N==1)
return 5;
else
return 5 + timesFive(N-1);
}
public static void main(String args[])
{
//Prompts the user to enter a nonnegative integer
int N = Integer.parseInt.(console.readLine("Enter a nonnegative integer: "));
//Outputs the sum, Bipower and TimesFive
System.out.println(sum(n));
System.out.println(BiPower(n));
System.out.println(TimesFive(n));
}
}
Answer:
The correct option is A
Explanation:
In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.
Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.
That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).
Cheers!
