We haven n! = (n-1)! x n and (n+1)! = n! x (n + 1);
Then, (n!)^2 = n! x n! = n! x (n-1)! x n;
And (n+1)!(n-1)! = n! x (n + 1) x (n-1)!;
Finally, [n! x (n-1)! x n] / [n! x (n + 1) x (n-1)!] = (n+1)/n;
APR is different than other compounding periods because you would need to find some equivelancy to compare things at.
5^5-9(200/4)-(10*90)/5-4^4(5)+156-256
= 3125-9( 200/4)-(10*90)/5-(4^4)(5)+156-256
= 3125-(9)(50)- (10*90)/5-(4^4)(5)+156-256
= 3125-450- (10*90)/5-(4^4)(5)+156-256
= 2675-(10*90)/5-(4^4)(5)+156-256
= 2675 - 900/5 - (4^4)(5)+156-256
= 2675 - 180 - (4^4)(5)+156-256
= 2495 - (4^4)(5)+156-256
= 2495 - 1280 + 156 -256
= 1215 + 156 - 256
= 1371 - 256
= 1115
I hope that's help , please if you have question(s) just let me know !
3a + a - 15 = 225 . your substituting b(a-15) into the first equation. <span />
Answer:
Hi how are you doing today Jasmine
