Question

The function prototype double mySqrt(int x); Declares a function called mySqrt which takes a double as an argument and returns an integer. Defines a function called mySqrt which takes an argument of type x and returns a double. Declares a function called mySqrt which takes an integer as an argument and returns a double. Defines a function called double which calculates square roots.

Answer 1

The prototype 'double mySqrt(int x);' declares a function called mySqrt which takes an integer as an argument and returns a double, most likely intended to calculate square roots.

Step-by-step explanation:

The function prototype double mySqrt(int x) does not declare a function that takes a double as an argument and returns an integer. Nor does it define a function called double which calculates square roots. Instead, this prototype declares a function named mySqrt that takes an integer as an argument and returns a double. This implies that the function mySqrt is likely designed to calculate the square root of an integer and return the result as a double-precision floating-point number.

Answer 2

A) declares a function called mySqrt that takes integer as argument and returns double.

Step-by-step explanation: