Preview, in MS word. It goes for print or publishing, whatever you desire.
Using e-mail to send messages is the best choice to convey urgent and highly sensitive information. E-mail is just a conversation between you and the recipient. So it is the best when it comes to when you are sending a highly sensitive information. While telephone fax letter and dispatch radio may need to use a mediator to transfer messages which violates the confidentiality of the information.
Answer:
Answered below
Explanation:
Some variable naming conventions include;
1) Variable should begin with either a letter or an underscore.
2) Variables having multiple words should have the first letter of every word after the first word, capitalized. This is the camelCase style.
3) variables should not be named after any of the inbuilt keywords except on special operations to override the original function of such keyword.
4) variable names are case-sensitive.
The importance of following these conventions is to maintain readability and consistency of code. Failure to follow these conventions may lead to chaotic codes, bugs and inefficient performance.
Answer:
length = float(input("Enter length of the backyard in foot: "))
width = float(input("Enter width of the backyard in foot: "))
sod_price = float(input("Enter the price of sod per square foot: "))
fencing_price = float(input("Enter the price of fencing per foot: "))
area = length * width
perimeter = 2 * (length + width)
cost = sod_price * area + fencing_price * perimeter
print("The cost of landscaping is $" + str(cost))
Explanation:
*The code is in Python.
Ask the user to enter the length, width, sod_price, and fencing_price
Calculate the area and perimeter of the backyard
Calculate the cost, sod_price * area + fencing_price * perimeter
Print the cost
/*
Since we have to check the first two options only as mentioned in last part of question the loop will work 2 times only and will compare the cost of first element and second and assign the healthoption accordingly
*/
for(int i =0;i<=1;i++){
if(annualCost[i]<annualCost[i+1]
best2 = healthOption[i]
else
best2 = healthOption[i+1]
}