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Inga [223]
3 years ago
14

4. Where do you find the command to create bibliography citations or a table of contents?

Computers and Technology
2 answers:
Mrac [35]3 years ago
4 0

Answer:

the answer is quick access toolbar

Vitek1552 [10]3 years ago
3 0
Answer is C. Quick access toolbar.
You might be interested in
What is a feature of webmail
Fantom [35]

Answer:

Webmail allows the users to access their emails as long as they have access to an Internet connection and a web browser. This also means that the user cannot read an old email or draft a new email offline.

8 0
3 years ago
Samantha is in the beginning stages of OOP program development. What are the five steps she must follow for creating an OOP prog
lora16 [44]

Answer:

1. Classes and objects

2. Inheritance

3. Polymorphism

4. Data hiding/ encapsulation

5. Interfaces.

Explanation:

Classes and objects depict the major component of the OOP (object oriented programming). It explains the object like a ball in a soccer game development.

The inheritance is like the subclass of the object. Data hiding is a stage in oop where the codes or data are hidden from another users.

In the polymorphism stage, the object is given the ability to change to a sub-object, while in the interface stage a function or method signature is defined without implementing it.

3 0
3 years ago
Read 2 more answers
Select the correct answer from each drop-down menu. Rita runs a small business that designs custom furnishings for corporate cli
LUCKY_DIMON [66]

Software as a Service cloud model is ideal for Rita’s business. SaaS are office solutions that allow Rita’s small business to work more efficiently and in a more organized way. Most SaaS applications are used for invoicing and accounting, sales, performance monitoring, and overall planning. SaaS applications can save Rita money. They do not require the deployment of a large infrastructure at her location. As a result, it drastically reduces the upfront commitment of resources. Whoever manages SaaS’s IT infrastructure running the applications brings down fees for software and hardware maintenance. SaaS has generally been acknowledged to be safer than most on-premise software.

5 0
3 years ago
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We informally define the term corresponding element as follows: The first element in an array and the last element of the array
yanalaym [24]

The corresponding element in an array can be written as follows.

The given array is arr[len] whose size is given by integer variable len. We take another integer variable k.

Corresponding element for arr[k] = arr[len-1-k]

This applies for all expressions of k beginning from 0 upto half the size of len, i.e., len/2.

For element, arr[0], the corresponding element is arr[10-1-0] = arr[9].

The above expression is valid if the sum of index of both element and its corresponding element adds upto the index of the last element of the array.

Example: k + (len – 1 - k)  

= k + len – 1 - k

= len – 1

= index of the last element of the array

The number of elements that an array can hold is one more than the index of the last element of the array.

Suppose an array of 10 elements is given. The above expression holds true since first element of the array is always written as arr[0] and last element will be written as arr[9] for an array of 10 elements.

This is illustrated below.

The c++ program to display the element and its corresponding element along with their index is shown below.

#include <iostream>

using namespace std;

int main() {    

   int len=10, arr[len],k;    

   for(k=0; k<len; k++)

   {

       // elements are initialized to twice their index

       arr[k] = k*2;

   }

   cout<<"Element "<<" Index "<<"\t"<<" Corresponding "<<" Index"<<endl;

   for(k=0; k<len/2; k++)

   {

       cout<<arr[k]<<"\t"<<"\t"<<"\t"<<k<<"\t"<<"\t"<<"\t"<<arr[len-1-k]<<"\t"<<"\t"<<len-1-k<<endl;

   }

}

OUTPUT

Element Index    Corresponding   Index

0  0   18  9

2  1   16  8

4  2   14  7

6  3   12  6

8  4   10  5

 

The above program initializes all the elements of the array to twice the value of their corresponding index. Following this, the element, its index, the corresponding element and its index is displayed.

All the above is achieved using for loop. The size of the array and its elements can be changed.

The expression for the corresponding element is thus explained.

7 0
3 years ago
I need the SQL statements for these questions:
zimovet [89]

Answer:

Explanation:

/* From the information provided, For now will consider the name of table as TRIPGUIDES*/

/*In all the answers below, the syntax is based on Oracle SQL. In case of usage of other database queries, answer may vary to some extent*/

1.

Select R.Reservation_ID, R.Trip_ID , C.Customer_Num,C.Last_Name from Reservation R, Customer C where C.Customer_Num=R.Customer_Num ORDER BY C.Last_Name

/*idea is to select the join the two tables by comparing customer_id field in two tables as it is the only field which is common and then print the desired result later ordering by last name to get the results in sorted order*/

2.

Select R.Reservation_ID, R.Trip_ID , R.NUM_PERSONS from Reservation R, Customer C where C.Customer_Num=R.Customer_Num and C.LAST_NAME='Goff' and C.FIRST_NAME='Ryan'

/*Here, the explaination will be similar to the first query. Choose the desired columns from the tables, and join the two tables by equating the common field

*/

3.

Select T.TRIP_NAME from TRIP T,GUIDE G,TRIPGUIDES TG where T.TRIP_ID=TG.TRIP_ID and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Abrams' and G.FIRST_NAME='Miles'

/*

Here,we choose three tables TRIP,GUIDE and TRIPGUIDES. Here we selected those trips where we have guides as Miles Abrms in the GUIDES table and equated Trip_id from TRIPGUIDES to TRIP.TRIP_Name so that can have the desired results

*/

4.

Select T.TRIP_NAME

from TRIP T,TRIPGUIDES TG ,G.GUIDE

where T.TRIP_ID=TG.TRIP_ID and T.TYPE='Biking' and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Boyers' and G.FIRST_NAME='Rita'

/*

In the above question, we first selected the trip name from trip table. To put the condition we first make sure that all the three tables are connected properly. In order to do so, we have equated Guide_nums in guide and tripguides. and also equated trip_id in tripguides and trip. Then we equated names from guide tables and type from trip table for the desired results.

*/

5.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='2016-07-23' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

/*

The explaination for this one will be equivalent to the previous question where we just equated the desired columns where we equiated the desired columns in respective fields and also equated the common entities like trip ids and customer ids so that can join tables properly

*/

/*The comparison of dates in SQL depends on the format in which they are stored. In the upper case if the

dates are stored in the format as YYYY-MM-DD, then the above query mentioned will work. In case dates are stored in the form of a string then the following query will work.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='7/23/2016' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

*/

6.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION R WHERE R.TRIP_ID IN

{SELECT TRIP_ID FROM TRIP T WHERE STATE='ME'}

/*

In the above question, we firstly extracted all the trip id's which are having locations as maine. Now we have the list of all the trip_id's that have location maine. Now we just need to extract the reservation ids for the same which can be trivally done by simply using the in clause stating print all the tuples whose id's are there in the list of inner query. Remember, IN always checks in the set of values.

*/

7.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION WHERE

EXISTS {SELECT TRIP_ID FROM TRIP T WHERE STATE='ME' and R.TRIP_ID=T.TRIP_ID}

/*

Unlike IN, Exist returns either true or false based on existance of any tuple in the condition provided. In the question above, firstly we checked for the possibilities if there is a trip in state ME and TRIP_IDs are common. Then we selected reservation ID, trip ID and Trip dates for all queries that returns true for inner query

*/

8.

SELECT G.LAST_NAME,G.FIRST_NAME FROM GUIDE WHERE G.GUIDE_NUM IN

{

SELECT DISTINCT TG.GUIDE_NUM FROM TRIPGUIDES TG WHERE TG.TRIPID IN {

SELECT T.TRIP_ID FROM TRIP T WHERE T.TYPE='Paddling'

}

}

/*

We have used here double nested IN queries. Firstly we selected all the trips which had paddling type (from the inner most queries). Using the same, we get the list of guides,(basically got the list of guide_numbers) of all the guides eds which were on trips with trip id we got from the inner most queries. Now that we have all the guide_Nums that were on trip with type paddling, we can simply use the query select last name and first name of all the guides which are having guide nums in the list returned by middle query.

*/

4 0
3 years ago
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