Answer:
The architectural pattern i will use for the school management is the client-server pattern.
This pattern would consist of a server and many clients. wherein the server component would provide services to that of the clients and its components as specified and also there would be a client request service from that of the server.
Explanation:
Solution
A school management system would always involve the client server pattern as this pattern would have a server and many clients wherein the server component would give services to that of the clients and its components as specified and also there would be a client request service from that of the server. This server would share the appropriate services to such clients and also listen to the client's requests.
Such kind of pattern would mostly be used for for the online platforms or application like that of document.
Answer:
//Program was implemented using C++ Programming Language
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
unsigned int second_a(unsigned int n)
{
int r,sum=0,temp;
int first;
for(int i= 1; I<=n; i++)
{
first = n;
//Check if first digit is 3
// Remove last digit from number till only one digit is left
while(first >= 10)
{
first = first / 10;
}
if(first == 3) // if first digit is 3
{
//Check if n is palindrome
temp=n; // save the value of n in a temporary Variable
while(n>0)
{
r=n%10; //getting remainder
sum=(sum*10)+r;
n=n/10;
}
if(temp==sum)
cout<<n<<" is a palindrome";
else
cout<<n<<" is not a palindrome";
}
}
}
Explanation:
The above code segments is a functional program that checks if a number that starts with digit 3 is Palindromic or not.
The program was coded using C++ programming language.
The main method of the program is omitted.
Comments were used for explanatory purpose.
Answer:
the elevation at point X is 2152.72 ft
Explanation:
given data
elev = 2156.77 ft
BS = 2.67 ft
FS = 6.72 ft
solution
first we get here height of instrument that is
H.I = elev + BS ..............1
put here value
H.I = 2156.77 ft + 2.67 ft
H.I = 2159.44 ft
and
Elevation at point (x) will be
point (x) = H.I - FS .............2
point (x) = 2159.44 ft - 6.72 ft
point (x) = 2152.72 ft
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
