Answer:
C. deriving benefits from highly focused and high technology markets
Explanation:
Firms that engage in cost-leadership approach seek to combine a low per-unit income with large sales for profit making purposes. Typically, but not always, they are more inclined to market their products and services to a large population base or a niche with a high demand volume. While differentiation enables a company to accomplish a competitive advantage. A Competitive advantage enables a company to achieve more strides over other companies offering related product substitutes. It is an important marketing process that is of critical economic importance to a business.
It should be noted that deriving benefits from highly focused and high technology markets is not part of the approaches to combining overall cost leadership and differentiation competitive advantages.
Answer:
In a <u>little endian computer</u> -The data's least substantial byte is put at the lower address byte. The remaining information will be put in memory in order in the next three bytes.
a)1234
4 is placed at the least significant bits,so this byte stored at lower memory address.
1 is placed at the most significant bits,so this byte stored at higher memory address.
b) ABFC
C is placed at the least significant bits,so this byte stored at lower memory address.
A is placed at the most significant bits,so this byte stored at higher memory address.
c) B100
0 is placed at the least significant bits,so this byte stored at lower memory address.
B is placed at the most significant bits,so this byte stored at higher memory address.
d) B800
0 is placed at the least significant bits,so this byte stored at lower memory address.
B is placed at the most significant bits,so this byte stored at higher memory address.
<span> is the person in charge of actually shooting the film. He is the head of the camera and lighting departments, and as such he has a big </span>role<span> in the making of any movie
</span>
let's say they catch-up to each other after t hour.
so
in t hour distance travelled by Renee
d = speed × time = 50t
in t hour distance travelled by Kim
d = speed × time = 60(t-1) + 0×1 = 60(t-1)
Note: here kim didn't covered any distance in first hour and in rest t-1 hour it travelled all distance
now.as distance travelled by both is same so
60(t-1) = 50t
60t -60 = 50t
adding 60 both sides
60t = 50t+60
subtracting 50t both sides
60t -50t = 50t+60 - 50t
10t = 60
dividing by 10 both sides
t = 60/10 = 6
so in 6 hour both will catch-up to each other
Answer:
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// function to compute sum and product
void square(int input)
{
// variable to store the square of number
long long squ_num;
calculate the square of number
squ_num=pow(input,2);
// print the output
cout<<"square of "<< input<<" is: "<<squ_num<<endl;
}
// driver function
int main()
{
int n;
// read the number until user enter 0
do{
cout<<"enter a number!! (0 to stop):";
// read the input from user
cin>>n;
// call the function to calculate square of input number
square(n);
}while(n!=0);
return 0;
}
Explanation:
Declare a variable "n" to read the input number from user.Call the function square() with parameter "n".In this function, it will calculate the square of the input number and print it.This will repeat until user enter 0 as input.Program will ends when user give 0 as input.
Output:
enter a number!! (0 to stop):5
square of 5 is: 25
enter a number!! (0 to stop):7
square of 7 is: 49
enter a number!! (0 to stop):11
square of 11 is: 121
enter a number!! (0 to stop):0
