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Oxana [17]
3 years ago
13

Which statement correctly describes one aspect of the team's commitment at the end of PI Planning?​

Computers and Technology
1 answer:
satela [25.4K]3 years ago
3 0

Explanation:

<h2><em><u>☆</u></em><em><u>T</u></em><em><u>H</u></em><em><u>A</u></em><em><u>N</u></em><em><u>K</u></em><em><u> </u></em><em><u>Y</u></em><em><u>O</u></em><em><u>U</u></em><em><u> </u></em><em><u>F</u></em><em><u>O</u></em><em><u>R</u></em><em><u> </u></em><em><u>T</u></em><em><u>H</u></em><em><u>E</u></em><em><u> </u></em><em><u>P</u></em><em><u>O</u></em><em><u>I</u></em><em><u>N</u></em><em><u>T</u></em><em><u>S</u></em><em><u>☆</u></em></h2>

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A _______ is a collection of software routines that can be used by other software. Licensing terms for this type of software are
tensa zangetsu [6.8K]

Answer: library

Explanation:

A library refers to the collection of software routines that can be used by other software. Licensing terms for this type of software are important for programmers who use the software.

It is the collection of non-volatile resources that is used by computer programs, usually for the development of software.

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3 years ago
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D. Click the Change button to accept the change.

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Please help with this coding question
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3 years ago
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5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0
3 years ago
What's wrong with these codes in code HS Karel challenges(7.1.2. Racing Karel) Codes: //Below is the program that have Karel mov
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Answer:

The program only runs 5 five since the for loop statement is limited to loop only five times.

Explanation:

In programming, a for-loop statement is used to repeat a collection of events a definite number of times. The number of loops is specified and compared with a variable to execute a block of code.

The for-loop statement in the code above declares and initializes a variable "i" to zero, runs the block of code, and increments by one if it is less than 5.

To make it run eight times, the value five should be changed to 8 instead.

5 0
3 years ago
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