IDNLearn.com offers a seamless experience for finding and sharing knowledge. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.