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leonid [27]
2 years ago
15

HELP ASAP U GET BRAINLIEST

Computers and Technology
2 answers:
hichkok12 [17]2 years ago
8 0

If you want to change the font to your slides, in the easiest way possible, just click the font option and choose the most suitable one, which is available on (B) tool bar.

The tool bar is generally pinned to the top of your Microsoft PowerPoint window, and contains multiple tabs. The tab with the font option is located in Home.

Korolek [52]2 years ago
7 0
B) <span>tool bar

i hope helped ^-^</span>
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Early Intel processors (e.g., the 8086) did not provide hardware support for dual-mode operation (i.e., support for a separate u
Firlakuza [10]

Answer:

Uneven use of resources

Explanation:

Potential problem associated with supporting multi - user operation without hardware support is:

Uneven use of resources: In a situation where we assign a set of resources to user 1 and if a new user comes, then it would be difficult to allocate new resources to him. The processor would get confused between the two users. And the tasks would not be completed. This can affect task processing.

8 0
2 years ago
Suppose your name was George Gershwin. Write a complete program that would print your last name, followed by a comma, followed b
NNADVOKAT [17]

Answer:

I will write the code in C++ and JAVA      

Explanation:

<h2>JAVA CODE</h2>

public class Main

{ public static void main(String[] args) {

       // displays Gershwin,George

     System.out.println("Gershwin,George"); }  }

<h2>C++ Code:</h2>

#include <iostream>

using namespace std;  

int main()

{  cout<<"Gershwin,George"; }    

// displays last name Gershwin followed by , followed by first name George

//displays Gershwin,George as output.

6 0
2 years ago
Read 2 more answers
Write a loop that displays all possible combinations of two letters where the letters are 'a', or 'b', or 'c', or 'd', or 'e'. T
Dafna11 [192]

Answer:

for (char outerChar='a'; outerChar<='e'; outerChar++){

for (char innerChar='a'; innerChar<='e'; innerChar++){

cout << outerChar << innerChar << "\n";

}

}

3 0
2 years ago
Consider the following relation:CAR_SALE(Car#, Date_sold, Salesperson#, Commission%, Discount_amt)Assume that a car may be sold
GalinKa [24]

Answer:

The answer to this question can be given as:

Normalization:

Normalization usually includes the division of a table into two or more tables as well as defining a relation between the table. It is also used to check the quality of the database design. In the normalization, we use three-level that are 1NF, 2NF, 3NF.

First Normal Form (1NF):

In the 1NF each table contains unique data. for example employee id.  

Second Normal Form (2NF):

In the 2NF form, every field in a table that is not a determiner of another field's contents must itself be a component of the table's other fields.

Third Normal Form (3NF):

In the 3NF form, no duplication of information is allowed.

Explanation:

The explanation of the question can be given as:

  • Since all attribute values are single atomic, the given relation CAR_SALE is in 1NF.  
  • Salesperson# → commission% …Given  Thus, it is not completely dependent on the primary key {Car#, Salesperson#}. Hence, it is not in 2 NF.                                                                                                        

The 2 NF decomposition:

        CAR_SALE_1(Car#, Salesperson#, Date_sold, Discount_amt)

         CAR_SALE_2(Salesperson#, Commission%)

  • The relationship in question is not in 3NF because the nature of a transitive dependence occurs
  • Discount_amt → Date_sold → (Car#, Salesperson#) . Thus, Date_sold is neither the key itself nor the Discount amt sub-set is a prime attribute.  

The 3 NF decomposition :

        CAR_SALES_1A(Car#, Salesperson#, Date_sold)

        CAR_SALES_1B(Date_sold, Discount_amt)

        CAR_SALE_3(Salesperson#, Commission%)

5 0
2 years ago
What would make this comparison statement False? Complete with the appropriate relational operator. "G" _____= "G"
My name is Ann [436]

Equality and Relational Operators

For the statement to return false, you can simply use the "not equal to" equality operation. The full symbol of this operation is '!=', disregarding the quotes.

<u>Examples:</u>

  • [1 != 1]  would produce FALSE. Translation: 1 <u>does not equal</u> 1?
  • [1 == 1]  would produce TRUE. Translation: 1 <u>does</u> 1?
  • ["G" != "G]  would produce <u>FALSE</u>. Translation: "G" <u>does not equal</u> "G"?

CONCLUSION: Use "!=".

8 0
2 years ago
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