no=3456 for x in reversed(str(no)): print(x) If you convert the number to a string and reverse it, you will obtain the same outcome.
Answer:
Below is the explanation for the C code.
Explanation:
#include <stdio.h>
#include <stdbool.h>
int main(void) {
int userNum;
bool isPositive;
bool isEven;
scanf("%d", &userNum);
isPositive = (userNum > 0);
isEven = ((userNum % 2) == 0);
if(isPositive && isEven){
printf("Positive even number");
}
else if(isPositive &&!isEven){
printf("Positive number");
}
else{
printf("Not a positive number");
}
printf("\n");
return 0;
}
Additional resources required for the projects
Added time necessary for the project
Clarification:
In any project management scenario, there will naturally be unexpected changes and additional needs, hence to successfully complete a project, one must allocate more time and resources. It is advisable that, based on the project specifics, the end user should maintain a sufficient buffer to accommodate any variations in human resources and the extra time necessary for project completion.
When planning the project, a consideration of extra time per each task is essential.
Every task within project management is categorized under distinct scopes of work.
Response:
Written in Python
G = 6.673 *pow(10,-11)
M = 5.98 *pow(10,24)
d = float(input("Enter distance: "))
g = (G * M)/(pow(d,2))
print("Calculated gravity acceleration: "+str(g))
Explanation:
The following line sets the gravitational constant
G = 6.673 *pow(10,-11)
This line establishes the mass of the Earth
M = 5.98 *pow(10,24)
This prompts the user to input the object's distance
d = float(input("Enter distance: "))
This calculates the gravity exerted on the object
g = (G * M)/(pow(d,2))
This line outputs the gravity acceleration without rounding
print("Calculated gravity acceleration: "+str(g))
Response: 1,500,000 bytes.
Clarification:
If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.
In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.
This indicates that we need to accommodate 12,000,000 bits.
Given that 1 byte equals 8 bits.
So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.
