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MrRa [10]
3 years ago
13

• R7.9 Write enhanced for loops for the following tasks.

Computers and Technology
1 answer:
alexdok [17]3 years ago
3 0

Answer:

a is the correct answer

Explanation:

correct me if I'm wrong hope it's help thanks

You might be interested in
Construct a class that will model a quadratic expression (ax^2 + bx + c). In addition to a constructor creating a quadratic expr
stich3 [128]

Answer:

Following are the code to this question:

#include <iostream>//header file

#include<math.h>//header file

using namespace std;

class Quadratic//defining a class Quadratic  

{

private:

double a,b,c;//defining a double variable

public:

Quadratic()//defining default constructor

{

a = 0;//assigning value 0  

b = 0;//assigning value 0  

c = 0;//assigning value 0  

}

Quadratic(double a, double b, double c)//defining a parameterized constructor  

{

this->a = a;//use this keyword to hold value in a variable

this->b = b;//use this keyword to hold value in b variable

this->c = c;//use this keyword to hold value in c variable

}

double getA() //defining a get method  

{

return a;//return value a

}

void setA(double a)//defining a set method to hold value in parameter

{

this->a = a;//assigning value in a variable

}

double getB() //defining a get method  

{

return b;//return value b

}

void setB(double b)//defining a set method to hold value in parameter  

{

this->b = b;//assigning value in b variable

}

double getC() //defining a get method

{

return c;//return value c

}

void setC(double c)//defining a set method to hold value in parameter

{

this->c = c;//assigning value in c variable

}

double Evaluate(double x)//defining a method Evaluate to hold value in parameter

{

return ((a*x*x)+(b*x)+c);//return evaluated value

}

double numberOfReal()//defining a method numberOfReal to calculates the real roots

{

return (b*b)-(4*a*c);//return real roots

}

void findroots()//defining a method findroots

{

double d=numberOfReal();//defining double variable to hold numberOfReal method value

if(d<0)//use if block to check value of d less than 0

cout<<"Equation has no real roots"<<endl;//print message

else

{

double r1=(-b+sqrt(numberOfReal()))/(2*a);//holding root value r1

double r2=(-b-sqrt(numberOfReal()))/(2*a);//holding root value r2

if(r1==r2)//defining if block to check r1 equal to r2

cout<<"Equation has one real root that is "<<r1<<endl;//print message with value

else//else block

cout<<"The equation has two real roots that are "<<r1<<" and "<<r2<<endl;////print message with value

}

}

void print()//defining a method print  

{

cout<< a << "x^2 + " << b << "x + " << c <<endl;//print Quadratic equation

}

};

int main()//defining main method  

{

Quadratic q(5,6,1);//creating Quadratic class object that calls parameterized constructor

q.print();//calling print method

cout<<q.numberOfReal()<<endl;//calling method numberOfReal that prints its value

q.findroots();//calling method findroots

cout<<q.Evaluate(-1);//calling method Evaluate that prints its value

return 0;

}

Output:

5x^2 + 6x + 1

16

The equation has two real roots that are -0.2 and -1

0

Explanation:

In the above code, a class "Quadratic" is declared, which is used to define a default and parameter constructor to holds its parameter value.

In the next step, the get and set method is defined that holds and returns the quadratic value, and "Evaluate, numberOfReal, findroots, and print" in the evaluate method a double variable is used as a parameter that returns evaluated value.

In the "numberOfReal" method it calculates the real roots and returns its value. In the "findroots" method a double variable "d" is declared that hold "numberOfReal" value,

and use a conditional statement to check its value, and in the print method, it prints the quadratic equation.

In the main method, the lass object it calls the parameterized constructor and other methods.

5 0
3 years ago
Suppose you have a file of data of approximately 100,000 personnel records (FYI, 213 = 8192 and 214 = 16384) using Social Securi
Natali [406]

Answer:

See explaination

Explanation:

The reasons why indexed sequential search structure is better are:

1. In index sequential search any field of the records can be used as the key. This key field can be numerical or alphanumerical.

2. Since each record has its data block address, searching for a record in larger database is easy and quick. There is no extra effort to search records. But proper primary key has to be selected to make efficient.

3. This method gives flexibility of using any column as key field and index will be generated based on that. In addition to the primary key and its index, we can have index generated for other fields too. Hence searching becomes more efficient, if there is search based on columns other than primary key.

The reasons why 5 B-tree is better:

1.The B-tree Provides support for range of queries in an efficient manner and You can iterate over an ordered list of elements.

2. B-Tree algorithms are good for accessing pages (or blocks) of stored information which are then copied into main memory for processing. In the worst case, they are designed to do dynamic set operations in O(lg n) time because of their high "branching factor" (think hundreds or thousands of keys off of any node). It is this branching factor that makes B-Trees so efficient for block storage/retrieval, since a large branching factor greatly reduces the height of the tree and thus the number of disk accesses needed to find any key.

3. It is a generalization of a BST in that a node can have more than two children. These are self-balancing and hence the average and worst complexities is logarithmic. We opt for these when the data is too huge to fit in main memory. These structures are used in database indexing and help in faster operations on disk

the additional informations is we should first decide to choose which structure is suiatable for which algoritm in terms of space and time and then use the appropriate search algorithm.

8 0
3 years ago
Which statement best describes a computer program?
ziro4ka [17]

Answer:

B is the best answer.

Explanation:

All other options listed are related to a program but B beast answers the question.

7 0
2 years ago
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest va
Ahat [919]

Answer:

Explanation:

From the given information:

a_{n-1} , a_{n-2}...a_o in binary is:

a_{n-1}\times 2^{n-1}  + a_{n-2}}\times 2^{n-2}+ ...+a_o

So, the largest number posses all a_{n-1} , a_{n-2}...a_o  nonzero, however, the smallest number has a_{n-2} , a_{n-3}...a_o all zero.

∴

The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times

i.e.

(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}

= \dfrac{1(2^n-1)}{2-1}

\mathbf{=2^n -1}

(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}

\mathbf {= 2 ^{n-1}}

Hence, the smallest value is \mathbf{2^{n-1}} and the largest value is \mathbf{2^{n}-1}

3 0
3 years ago
Adam is so good at playing arcade games that he will win at every game he plays. One fine day as he was walking on the street, h
ohaa [14]

Answer:

ask it to ur teacher boiiiiii

3 0
3 years ago
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