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slavikrds [6]
3 years ago
11

20. Carmen can buy bottles of paint for $2.00 each and boxes of colored pencils for $3.50 each. She can spend no more than

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

(a) The equality that express many bottles of paint, x, and boxes of colored pencils, y, Carmen can buy is  2 x + 3.5 y = 42

b)Three different solutions:

x  = 14, y = 4 is First Solution.

x  = 7, y = 8 is Second Solution.

x  = 21, y = 0 is Third Solution.

Step-by-step explanation:

Here, the cost of 1 bottle of paint = $2.00 each

The cost of 1 box of colored pencils  = $3.5 each

Let us assume the number of bottle of paints purchased = x

So, the cost of x bottle of paints   = x ( Cost of 1 bottle of paint)

= x ($2.00)  = 2 x

Also, assume the number of box of colored pencils purchased = y

So, the cost of y box of colored pencils = y ( Cost of 1 box)

= y ($3.50)  = 3.5 y

Also, the total amount to be spent on art supplies  = $42

So, the total amount spent on x paint bottles + y box of colored pencil

= $42

or,  2 x + 3.5 y = 42

a ) So, the equality that express many bottles of paint, x, and boxes of colored pencils, y, Carmen can buy is  2 x + 3.5 y = 42

b)Three different solutions:

When y = 4 ,  equation is:   2 x + 3.5(4)  = 42

or,  2 x = 42 - 14  = 28, or x = 28/2  =  14

So, x  = 14, y = 4 is First Solution.

When y = 8 ,  equation is:   2 x + 3.5(8)  = 42

or,  2 x = 42 - 28  = 14, or x = 14/2  =  7

So, x  = 7, y = 8 is Second Solution.

When y = 0 ,  equation is:   2 x + 3.5(0)  = 42

or,  2 x = 42 , or x = 42/2  = 21

So, x  = 21, y = 0 is Third Solution.

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3 years ago
Im confused, can someone help?
Bumek [7]
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3 years ago
Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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