Answer:
Explanation:
% Clears variables and screen
clear; clc
% Asks user for input
n = input('Total number of objects: ');
r = input('Size of subgroup: ');
% Computes and displays permutation according to basic formulas
p = 1;
for i = n - r + 1 : n
p = p*i;
end
str1 = [num2str(p) ' permutations'];
disp(str1)
% Computes and displays combinations according to basic formulas
str2 = [num2str(p/factorial(r)) ' combinations'];
disp(str2)
=================================================================================
Example: check
How many permutations and combinations can be made of the 15 alphabets, taking four at a time?
The answer is:
32760 permutations
1365 combinations
==================================================================================
Express it in standard form and apply the basic indices laws to simplify
Algorithm of the Nios II assembly program.
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
and
The decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
<h3>The Algorithm and
decimal equivalent on the
seven-segment displays HEX3-0</h3>
Generally, the program will be written using a cpulator simulator in order to attain best result.
We are to
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
This will be the Algorithm of the Nios II assembly program .
Hence, the decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
For more information on Algorithm
brainly.com/question/11623795
Answer:
If a truss buckles or overturns, it is usually because of the failure of an adjacent truss or its bracing. A steel truss in a fire may buckle and overturn because of expansion or weakening from the heat. Most truss failures are the result of broken connections. Photo 1 shows a set of parallel-chord wood trusses supporting a plywood floor deck.
Explanation:
Answer:
slenderness ratio = 147.8
buckling load = 13.62 kips
Explanation:
Given data:
outside diameter is 3.50 inc
wall thickness 0.30 inc
length of column is 14 ft
E = 10,000 ksi
moment of inertia 
Area 
r = 1.136 in
slenderness ratio 
buckling load 
