Answer:I have no clue if you find out let me know
Explanation:
Answer:
The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.
Query:
SELECT ReservationlD, Trip.TripName. Customer.LastName. Customer.FirstName. (TripPrice+OtherFees) 'NumPersons as TotalCost FROM Reservation, Trip, Customer WHERE NumPersons>4 AND Reservation.TriplD=Trip.TriplD AND
Customer. CustomerNum=Reservation.CustomerNum:
Explanation:
- Select clause is used to retrieve data from specified database table or relation and returns the data in the form of table.
- ReservationID. Trip.TripName. Customer.LastName. Customer.FirstName are the column name of table.
- (TripPrice+OtherFees) 'NumPersons will calculate the total cost of the Trip and stored it into TotalCost column.
- As clause is used to give new name TotalCost to resultant column.
- FROM clause specifies one or more table from where records to be retrieved. o Reservation. Trip and Customer are the table name.
- WHERE clause is used in SQL query to retrieve only those records that satisfy the specified condition
Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K =
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators