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)
√[(5-1)² + (5-5)²]
√[16+0]
√(16)
+4 or -4 units
I hope this helped :D
Answer:
This is an Invalid question to answer. Please edit your question so people can answer it.
50000 - 5000 ten times and then youll end up with 0
The car wouldnt be worth anything
Answer:
2.5
Step-by-step explanation:
4 x _ = 10
_ = 10/4
_ = 2.5
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