Question

colton is painting a set of figurines a solid color. it takes him 15 minutes and 5 milliliters of paint to finish a small figurine. it takes him 30 minutes and 8 milliliters of paint to complete a large figurine. colton plans to paint for at least 360 minutes, and he can use a maximum of 100 milliliters of paint. this graph and system of inequalities represent this situation, where x represents the number of completed small figurines and y represents the number of completed large figurines.
$15x + 30y \geq 360$
$5x + 8y \leq 100$

Explanation:

Since the problem is not fully stated (it seems to be a problem about linear inequalities and feasible regions, maybe asking which point is in the feasible region or something similar, but the question part is missing), we can't provide a solution. However, if we assume it's a common problem like checking if a point \((x,y)\) satisfies the inequalities, here's how we would do it:

Step-by-Step Format (assuming we need to check a point, say \((x,y)=(10,10)\)):

Step1: Check the first inequality \(15x + 30y \geq 360\)

Substitute \(x = 10\) and \(y = 10\) into the left - hand side of the first inequality:
\(15\times10+30\times10=150 + 300=450\)
Since \(450\geq360\), the point satisfies the first inequality.

Step2: Check the second inequality \(5x + 8y \leq 100\)

Substitute \(x = 10\) and \(y = 10\) into the left - hand side of the second inequality:
\(5\times10+8\times10=50 + 80 = 130\)
Since \(130>100\), the point \((10,10)\) does not satisfy the second inequality.

If you can provide the complete question (e.g., which of the following points is a solution to the system of inequalities, or find the maximum number of figurines, etc.), I can give a more accurate answer.

Answer:

Step1: Check the first inequality \(15x + 30y \geq 360\)

Substitute \(x = 10\) and \(y = 10\) into the left - hand side of the first inequality:
\(15\times10+30\times10=150 + 300=450\)
Since \(450\geq360\), the point satisfies the first inequality.

Step2: Check the second inequality \(5x + 8y \leq 100\)

Substitute \(x = 10\) and \(y = 10\) into the left - hand side of the second inequality:
\(5\times10+8\times10=50 + 80 = 130\)
Since \(130>100\), the point \((10,10)\) does not satisfy the second inequality.

If you can provide the complete question (e.g., which of the following points is a solution to the system of inequalities, or find the maximum number of figurines, etc.), I can give a more accurate answer.