Answer:
A and C are the only <u>legal</u> but unethical options
Smaller down and monthly payments than leasing a car
Option A.
<u>Explanation:</u>
If you lease a car, then the amount of money that a person has to pay in the form of monthly payments and installments is less when compared to when a person owns a car.
When you own a car, the monthly payment made is more than the monthly payments compared to when the car has to be leased. This is the disadvantage of owning a car, that the payment is more to be made.
Answer:
public static String repeat(String text, int repeatCount) {
if(repeatCount < 0) {
throw new IllegalArgumentException("repeat count should be either 0 or a positive value");
}
if(repeatCount == 0) {
return "";
} else {
return text + repeat(text, repeatCount-1);
}
}
Explanation:
Here repeatCount is an int value.
at first we will check if repeatCount is non negative number and if it is code will throw exception.
If the value is 0 then we will return ""
If the value is >0 then recursive function is called again untill the repeatCount value is 0.
Answer:
This solution is implemented in C++
void makePositive(int arr[]){
int i =0;
while(arr[i]!=0){
if(arr[i]<0){
arr[i] = abs(arr[i]);
}
i++;
}
i = 0;
while(arr[i]!=0){
cout<<arr[i]<<" ";
i++;
}
}
Explanation:
This defines the function makePositive
void makePositive(int arr[]){
This declares and initializes i to 0
int i =0;
The following iteration is repeated until the last element in the array
while(arr[i]!=0){
This checks if current array element is negative
if(arr[i]<0){
If yes, it changes it to positive
arr[i] = abs(arr[i]);
}
The next element is then selected
i++;
}
This sets i to 0
i = 0;
The following iteration prints the updated content of the array
<em> while(arr[i]!=0){
</em>
<em> cout<<arr[i]<<" ";
</em>
<em> i++;
</em>
<em> }
</em>
}
See attachment for full program which includes the main
