It should be 0.3181818181818 so it would be a repeating decimal
They give us 2 pieces to the puzzle. Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106. (x^2 + y^2 = 106).
substitute the value for x into the second equation.
(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106 (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106 (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0 (factor)
(y-5) (5y-21) = 0 (set to 0)
y - 5 = 0
y = 5
substitute the 5 into the equation for y (x = 2(5) - 1)
x = 9 if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.
our answers are 5 and 9.
The answer is 3 + 2k
This is because multiplying by 1/5 is like dividing each term by 5
10k + 15 = 5 • (2k + 3)
<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:
A 0.01
Step-by-step explanation:
<em>I saw the answer on the internet.</em>
