Answer:
(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
Explanation:
Solution
(a) Let Mn denotes the number of maintenance visits after the nth journey
Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3
We therefore, notice that M follows an arithmetic sequence
So,
Mn = M₁ + (n-1) (M₂ -M₁)
= 1 + (n- 1) 1 = n
or Mn =n
(b) For what value of n we will get fro Mn > 10
Thus,
n > 10 (exceed 10) or n =11
(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused
So,
n >50 or n= 51
After making a journey of 51 times, the rocket will be discarded
Answer:
maneuverability
Explanation:
needless to say, I took the quiz
Da, sigur. cu ce ai nevoie de ajutor?
Answer:
/* C Program to rotate matrix by 90 degrees */
#include<stdio.h>
int main()
{
int matrix[100][100];
int m,n,i,j;
printf("Enter row and columns of matrix: ");
scanf("%d%d",&m,&n);
/* Enter m*n array elements */
printf("Enter matrix elements: \n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&matrix[i][j]);
}
}
/* matrix after the 90 degrees rotation */
printf("Matrix after 90 degrees roration \n");
for(i=0;i<n;i++)
{
for(j=m-1;j>=0;j--)
{
printf("%d ",matrix[j][i]);
}
printf("\n");
}
return 0;
}
Answer: 8.33333333 or 6.1989778
Explanation:
