Answer:
Boom operator- Microphone
Director of photography- video camera
Art Director- lighting
Gaffer- set building
Answer:
The correct option is communication diagram
Explanation:
The communication diagram represents the change of values for an item by a system.
A communication diagram is an expansion of the diagram of objects showing the objects together with the texts traveling from one to another. Besides the connections between objects, the communication diagram demonstrates the messages that the objects send to one another.
The correct option is a) communication diagram
ASDF JKL; are the home row keys.
1.
name = input("Enter your name: ")
num1 = int(input("Hello "+name+ ", enter an integer: "))
num2 = int(input(name+", enter another integer: "))
try:
if num1 % num2 == 0:
print("The first number is divisible by the second number")
else:
print("The first number is not divisible by the second number")
except ZeroDivisionError:
print("The first number is not divisible by the second number")
try:
if num2 % num1 == 0:
print("The second number is divisible by the first number")
else:
print("The second number is not divisible by the first number")
except ZeroDivisionError:
print("The second number is not divisible by the first number")
2.
import random, math
num1 = float(input("Enter a small decimal number: "))
num2 = float(input("Enter a large decimal number: "))
r = round(random.uniform(num1, num2), 2)
print("The volume of a sphere with radius " + str(r) + " is " + str(round(((4 / 3) * math.pi * (r ** 3)), 2)))
I hope this helps!
Answer:
The algorithm to find A is even or odd.
- input A.
- Check the remainder on diving by 2 by A%2.
- If remainder is 1 then A is odd Print(Odd).
- If remainder is 0 print(Even).
Explanation:
To check if the number is even or odd we use modulo operator(%).Which gives the remainder on dividing.So if we do this A%2 it will give the remainder that will come out on dividing the value of A by 2.
So if the remainder comes out is 1 then the number is odd and if the remainder is 0 then the number is odd.
