Nr of chocolates = c
nr of bags of 5 = a
nr of bags of 7 = b
50<c<70
c=5*a+2
c= 7*b-1 therefore b can be max 10 (if b=11 then c=76>70 can't do)
let's say b=10 then c=69 but in this case a=13.4 not integer (also can't do)
let's say b=9 then c=62 then a = 12 wich is ok
solution 1 c=62
let's also try for b=8 wich means c=55 and a=10.6, not integer, not ok again
and also for b=7 wich means c=48 but 48<50 so no go again and also for b<7 wich means c will grow smaller and smaller
so the only solution is c=62
Answer:
Step-by-step explanation:
This is because you use Pythagorean theorem. a^2 + b^2 = c^2.
Answer:
6457 + 7925 = 14382
From multiplying the binomial expression
(n+3)(n-9)
It would be n^2 - 9n + 3n - 27 or simplified
n^2 -6n - 27.
To evaluate the combination we proceed as follows:
21C3
Given nCk we shall have:
n!/[(n-k)!k!]
thus plugging the values in the expression we get:
21!/[(21-3)!3!]
=21!/(18!×3!)
=1330
Answer: 1330