You have exponential decay if the base is a between 0 and 1 and the exponent grows as x grows, or if the base is greater than 1 and the exponents goes to negative infinity as x grows.
The only function that satisfied these requests is the second one: 1/2 is between 0 and 1, and the exponent is x itself.
<u>the correct question is</u>
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.
Answer:
10.4% or 13/125
Step-by-step explanation:
5 perfect squares in the set of numbers
13 odd numbers
(5/25) x (13/25) = 0.104
0.104 x 100 = 10.4%
i hope its right
Let
. Then
, and so as
, you have
. The limit is then equivalent to
