Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Let's analyze each of the given statements based on the quadratic equation [tex]\( y = -6x^2 + 100x - 180 \)[/tex], where [tex]\( x \)[/tex] represents the selling price of each soccer ball and [tex]\( y \)[/tex] represents the daily profit.
1. If the store charges [tex]$\$[/tex] 5[tex]$ per soccer ball, it can expect to sell 170 of them. - To verify this, we substitute \( x = 5 \) into the quadratic equation to calculate the profit: \[ y = -6(5)^2 + 100(5) - 180 \] \[ y = -6(25) + 500 - 180 \] \[ y = -150 + 500 - 180 \] \[ y = 170 \] Thus, the profit when the store charges \$[/tex]5 per soccer ball is indeed \[tex]$170. 2. If the store charges $[/tex]\[tex]$ 8$[/tex] per soccer ball, it makes a daily profit of [tex]$\$[/tex] 236[tex]$. - To verify this, we substitute \( x = 8 \) into the quadratic equation to calculate the profit: \[ y = -6(8)^2 + 100(8) - 180 \] \[ y = -6(64) + 800 - 180 \] \[ y = -384 + 800 - 180 \] \[ y = 236 \] Thus, the profit when the store charges \$[/tex]8 per soccer ball is \[tex]$236. 3. If the store sells 12 soccer balls, it makes \$[/tex]156 in profit.
- This statement involves the number of soccer balls sold, but our original quadratic equation relates to the selling price, not the number of soccer balls sold. Assuming a misinterpretation:
[tex]\[ y = profit(12) \][/tex]
Substituting [tex]\( x = 12 \)[/tex] into the equation:
[tex]\[ y = -6(12)^2 + 100(12) - 180 \][/tex]
[tex]\[ y = -6(144) + 1200 - 180 \][/tex]
[tex]\[ y = -864 + 1200 - 180 \][/tex]
[tex]\[ y = 156 \][/tex]
Thus, if the store's equation is assumed based on pricing and not actual balls sold, it makes \[tex]$156 when considering evaluation for \( x=12 \). 4. The greater the selling price, the greater the profit. - To verify this, we consider the nature of the quadratic equation \( y = -6x^2 + 100x - 180 \): The quadratic term \(-6x^2\) indicates that the equation represents a parabola opening downwards. This means there is a maximum profit at a certain price, but not an ever-increasing profit: Substituting various reasonable prices will show the maximum and decreasing trend. Additionally, the Python output confirms this pattern is not increasing consistently: \[ greater\_profit\_check = False \] In summary, the true statements based on our calculations are: - If the store charges \$[/tex]5 per soccer ball, it can expect to make a profit of \[tex]$170. - If the store charges \$[/tex]8 per soccer ball, it makes a daily profit of \[tex]$236. - If evaluated at selling price $[/tex]12, the resulting profit calculation indicates \$156.
- The statement that greater prices always yield greater profits is false.
1. If the store charges [tex]$\$[/tex] 5[tex]$ per soccer ball, it can expect to sell 170 of them. - To verify this, we substitute \( x = 5 \) into the quadratic equation to calculate the profit: \[ y = -6(5)^2 + 100(5) - 180 \] \[ y = -6(25) + 500 - 180 \] \[ y = -150 + 500 - 180 \] \[ y = 170 \] Thus, the profit when the store charges \$[/tex]5 per soccer ball is indeed \[tex]$170. 2. If the store charges $[/tex]\[tex]$ 8$[/tex] per soccer ball, it makes a daily profit of [tex]$\$[/tex] 236[tex]$. - To verify this, we substitute \( x = 8 \) into the quadratic equation to calculate the profit: \[ y = -6(8)^2 + 100(8) - 180 \] \[ y = -6(64) + 800 - 180 \] \[ y = -384 + 800 - 180 \] \[ y = 236 \] Thus, the profit when the store charges \$[/tex]8 per soccer ball is \[tex]$236. 3. If the store sells 12 soccer balls, it makes \$[/tex]156 in profit.
- This statement involves the number of soccer balls sold, but our original quadratic equation relates to the selling price, not the number of soccer balls sold. Assuming a misinterpretation:
[tex]\[ y = profit(12) \][/tex]
Substituting [tex]\( x = 12 \)[/tex] into the equation:
[tex]\[ y = -6(12)^2 + 100(12) - 180 \][/tex]
[tex]\[ y = -6(144) + 1200 - 180 \][/tex]
[tex]\[ y = -864 + 1200 - 180 \][/tex]
[tex]\[ y = 156 \][/tex]
Thus, if the store's equation is assumed based on pricing and not actual balls sold, it makes \[tex]$156 when considering evaluation for \( x=12 \). 4. The greater the selling price, the greater the profit. - To verify this, we consider the nature of the quadratic equation \( y = -6x^2 + 100x - 180 \): The quadratic term \(-6x^2\) indicates that the equation represents a parabola opening downwards. This means there is a maximum profit at a certain price, but not an ever-increasing profit: Substituting various reasonable prices will show the maximum and decreasing trend. Additionally, the Python output confirms this pattern is not increasing consistently: \[ greater\_profit\_check = False \] In summary, the true statements based on our calculations are: - If the store charges \$[/tex]5 per soccer ball, it can expect to make a profit of \[tex]$170. - If the store charges \$[/tex]8 per soccer ball, it makes a daily profit of \[tex]$236. - If evaluated at selling price $[/tex]12, the resulting profit calculation indicates \$156.
- The statement that greater prices always yield greater profits is false.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.