def dx(fn, x, delta=0.001):
return (fn(x+delta) - fn(x))/delta
def solve(fn, value, x=0.5, maxtries=1000, maxerr=0.00001):
for tries in xrange(maxtries):
err = fn(x) - value
if abs(err) < maxerr:
return x
slope = dx(fn, x)
x -= err/slope
raise ValueError('no solution found')
Hi!
I have a slight understanding of excel, it depends on what you need.
Answer:
import java.util.Scanner;
public class num8 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter the mass");
double mass = in.nextDouble();
double weight = calWeight(mass);
System.out.println("The weigth is "+weight);
}
static double calWeight(double mass){
double weight = mass*9.80665; // assume a = accelation due to gravity = 9.80665N
if(weight>500){
System.out.println("Too Heavy");
}
else if(weight<100){
System.out.println("Too Light");
}
return weight;
}
}
Explanation:
- Using Java programming language
- The main method is created to request and store a variable, mass in kilogram. The main method call calWeight() and passes the value for mass
- A method calWeight() is created that calculates the weight in newtons (mass * 9.8).
- The method checks if the weight is greater than 500 (prints too heavy) if less than 100(prints to light)
- Returns the weight
Answer:
A. VHS
I took web design and past
