Question

Cindy won $50.00 for getting first place in the Science Fair. She spent her winnings on robotics kits and chemistry sets. Each robotics kit (y) costs $10.00, while each chemistry set costs $8.00 (x). Which of the following is a viable solution to the number of robotics kits and chemistry sets Cindy can purchase, assuming she spends her entire winnings?(1 point) Responses (0,5) left parenthesis 0 comma 5 right parenthesis (−5,9) left parenthesis negative 5 comma 9 right parenthesis (3,2.6) left parenthesis 3 comma 2.6 right parenthesis (5,1)

Answer 1

To find viable solutions for the number of robotics kits (y) and chemistry sets (x) Cindy can purchase, we can set up the following equation based on her winnings of $50.00:

\[ 10y + 8x = 50 \]

Now we'll evaluate each of the provided options:

1. (0, 5):

   • Substitute \(y = 0\) and \(x = 5\) into the equation: \[ 10(0) + 8(5) = 0 + 40 = 40 \quad (\text{Not valid, less than } 50) \]

2. (−5, 9):

   • Substitute \(y = -5\) and \(x = 9\) into the equation: \[ 10(-5) + 8(9) = -50 + 72 = 22 \quad (\text{Not valid, cannot buy a negative quantity}) \]

3. (3, 2.6):

   • Substitute \(y = 3\) and \(x = 2.6\) into the equation: \[ 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \quad (\text{Not valid, exceeds } 50) \]

4. (5, 1):

   • Substitute \(y = 5\) and \(x = 1\) into the equation: \[ 10(5) + 8(1) = 50 + 8 = 58 \quad (\text{Not valid, exceeds } 50) \]

Based on this analysis, none of the provided options calculate to a valid purchase that uses exactly $50. However, if we want to check which solution is closest or fits the criteria of values greater than or equal to $0 and spending exactly $50, we find that only non-listed pairs would satisfy spending exactly their winnings, while having values \(x \geq 0\) and \(y \geq 0\).

Thus, according to the provided options, none of them are valid.