9. What is composition? Why is composition important?
1 answer:
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Answers:
(3) True
(4) Num Lock
Explanations:
(3) To undo any task in most of the softwares including Blender, CTRL + Z is used. If you want to double check, go to preference -> Input -> Search for undo. You would see CTRL + Z shortcut there for one time undo (as shown in the picture 1 attached.) Hence (TRUE).
(4) As in blender, during the scene creation, the designers usually use top, left, right or bottom view. Blender has the built-in keys set for those views. As you can see in the second image that the Top view is set to NumPad 7 (likewise others set to other numbers). In order to use the those numpad numbers, you must turn on the NumLock! Hence NumLock is the answer.
(3) True
(4) Num Lock
Explanations:
(3) To undo any task in most of the softwares including Blender, CTRL + Z is used. If you want to double check, go to preference -> Input -> Search for undo. You would see CTRL + Z shortcut there for one time undo (as shown in the picture 1 attached.) Hence (TRUE).
(4) As in blender, during the scene creation, the designers usually use top, left, right or bottom view. Blender has the built-in keys set for those views. As you can see in the second image that the Top view is set to NumPad 7 (likewise others set to other numbers). In order to use the those numpad numbers, you must turn on the NumLock! Hence NumLock is the answer.
Answer:
Check the explanation
Explanation:
CODE:-
import java.util.*;
class UserName{
ArrayList<String> possibleNames;
UserName(String firstName, String lastName){
if(this.isValidName(firstName) && this.isValidName(lastName)){
possibleNames = new ArrayList<String>();
for(int i=1;i<firstName.length()+1;i++){
possibleNames.add(lastName+firstName.substring(0,i));
}
}else{
System.out.println("firstName and lastName must contain letters only.");
}
}
public boolean isUsed(String name, String[] arr){
for(int i=0;i<arr.length;i++){
if(name.equals(arr[i]))
return true;
}
return false;
}
public void setAvailableUserNames(String[] usedNames){
String[] names = new String[this.possibleNames.size()];
names = this.possibleNames.toArray(names);
for(int i=0;i<usedNames.length;i++){
if(isUsed(usedNames[i],names)){
int index = this.possibleNames.indexOf(usedNames[i]);
this.possibleNames.remove(index);
names = new String[this.possibleNames.size()];
names = this.possibleNames.toArray(names);
}
}
}
public boolean isValidName(String str){
if(str.length()==0) return false;
for(int i=0;i<str.length();i++){
if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))
return false;
}
return true;
}
public static void main(String[] args) {
UserName person1 = new UserName("john","smith");
System.out.println(person1.possibleNames);
String[] used = {"harta","hartm","harty"};
UserName person2 = new UserName("mary","hart");
System.out.println("possibleNames before removing: "+person2.possibleNames);
person2.setAvailableUserNames(used);
System.out.println("possibleNames after removing: "+person2.possibleNames);
}
}
Kindly check the attached image below for the code output.
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
