Complete question :
The denarius was a unit of currency in ancient Rome. Suppose it costs the Roman government 10 denarius per day to support 3 legionaries and 3archers. It only costs 3 denarius per day to support one legionary and one archer. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Answer:
No unique solution
Step-by-step explanation:
Given that:
Cost of supporting 3 legion aries and 3 archers = 10 denarius daily
Cost of supporting one legionary and one archer = 3 denarius
Let:
Legionaries = l ; archers = a
Equation for the first sentence :
3l + 3a = 10 - - - (1)
Second sentence :
l + a = 3 - - - - - (2)
From (2)
l = 3 - a
Substituting l = 3 - a into (1)
3(3 - a) + 3a = 10
9 - 3a + 3a = 10
9 - 0 = 10
The variable cancels out, Hence, ( there is no unique solution to find the cost of each soldier)
50% chance of a positive number. there are 30 numbers and 15 if them are positive. 15 is half of 30
Answer:
$16.45
Step-by-step explanation:
0.15 x 109.66= 16.45
Answer:
- Domain- set of all independent variables
- Range- set of all dependent variables
- Function- a relation where each value of the independent variable corresponds with only one value of the dependent variable
Step-by-step explanation:
20:28 plus 9 hours = 05:28
05:28 plus 10 minutes = 05:38
Answer= 05:38