Tan 25.5 = h/x
Tan 20.5 = h / x+1
x tan 25.5 = (x+1) tan 20.5
x 0.476975532 = (x+1) 0.373884679
0.476975532x = 0.373884679x + 0.373884679
Deduct both sides by .373884679
0.103090853x = 0.373884679
x = 3.6267 is the first person.
Add one to get for the second person
X = 4.6267
Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
The slope equals 0.
Hope this helps.
Given that the ratio of three trays is A : B : C = 2 : 3 : 4
Let us consider the common factor among the number of trays be "x".
If we need to make a total of 171 trays, then the equation of sum of trays would be :-
Type A + Type B + Type C = 171
2x + 3x + 4x = 171
9x = 171
x =
= 19
So we have following number of trays :-
Number of type A trays = 2x = 38 trays.
Number of type B trays = 3x = 57 trays.
Number of type C trays = 4x = 76 trays.
Now if each tray contains 20 cookies, then we would have following arrangements :-
Cookies in type A trays = 38×20 = 760 cookies.
Cookies in type B trays = 57×20 = 1140 cookies.
Cookies in type C trays = 76×20 = 1520 cookies.