Answer:
D) 11
Step-by-step explanation:
If you start at five and count back on each little x
than you should come up with 11
0=3
1=1
2=2
3=4
4=1
3+1+2+4+1=11
You do This because you need to find how many has fewer so you don't count 5
Answer:
IM SLOWWWWWWWWWWWWWWWWWWWWWw
Step-by-step explanation:
Answer:

Step-by-step explanation:
let
denote grams of
formed in
mins.
For
of
we have:
of A and
of B
Amounts of A,B remaining at any given time is expressed as:
of A and
of B
Rate at which C is formed satisfies:

Apply the initial condition,
,to the expression above

Now at
:

Substitute in X(t) to get
