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liberstina [14]
2 years ago
15

Angie and Kenny play online video games. Angie buys 1 software package and 1month of gameplay. Kenny buys 1 software package and

2 months of game play. Each software package costs ​$30. If their total cost is ​$90​, what is the cost of one month of gameplay?
Mathematics
1 answer:
noname [10]2 years ago
5 0

Answer: $10

Step-by-step explanation:

Angie buys = 1 software package and 1month of gameplay.

Kenny buys = 1 software package and 2 months of game play.

Cost of each software package = $30. Total cost = ​$90​

The total cost is the addition of what Kenny and Angie bought. This will be:

(1 SF + 1 G) + (1 SF + 2 G) = 90

where,

SF = software package = 30

G = game

(1 SF + 1 G) + (1 SF + 2 G) = 90

30 + 1G + 30 + 2G = 90

60 + 3G = 90

3G = 90 - 60

3G = 30

G = 30/3

G = 10

One month of game play cost $10

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Answer:

See explanation

Step-by-step explanation:

Let x be the number of spade shovels, y -the number of flat shovels and z - the number of square showels sold that day.

The store keeps an inventory of 80 shovels, then

x+y+z=80

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16x+9.60y+12.80z=1,072

a) The system of three equations is

\left\{\begin{array}{l}x+y+z=80\\ \\x=2z\\ \\16x+9.60y+12.80z=1,072\end{array}\right.

b) In matrix form this is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}80\\0\\1,072\end{array}\right)

c) The determinant is

\left\|\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right\|=0-32+9.60-0+19.20-12.80=-16

d) Find three determinants:

\left\|\begin{array}{ccc}80&1&1\\ 0&0&-2\\ 1,072&9.60&12.80\end{array}\right\|=0-2,144+0-0+1,536-0=-608

\left\|\begin{array}{ccc}1&80&1\\ 1&0&-2\\ 16&1,072&12.80\end{array}\right\|=0-2,560+1,072-0+2,144-1,024=-368

\left\|\begin{array}{ccc}1&1&80\\ 1&0&0\\ 16&9.60&1,072\end{array}\right\|=0+0+768-0-0-1,072=-304

So,

x=\dfrac{-608}{-16}=38\\ \\y=\dfrac{-368}{-16}=23\\ \\z=\dfrac{-304}{-16}=19

e) If the store doubled all prices and inventory, then the new matrix is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 32&19.20&25.60\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}160\\0\\2,144\end{array}\right)

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