Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Sure, let's compare the passes made by Ana and Taylor by analyzing the given information.
### Ana's Pass:
We have been given the vertical height [tex]\( a(x) \)[/tex] of Ana's pass at different values of [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 0 & 10 & 20 & 30 & 40 & 50 & 60 \\ \hline a(x) & 0 & 20 & 32 & 36 & 32 & 20 & 0 \\ \hline \end{array} \][/tex]
To find the maximum height of Ana's pass, we will look at the y-values in the table. The maximum height is [tex]\( 36 \)[/tex] feet.
### Taylor's Pass:
Taylor's pass is modeled by the quadratic equation:
[tex]\[ t(x) = -0.05(x^2 - 50x) \][/tex]
This equation can be expanded and simplified to:
[tex]\[ t(x) = -0.05x^2 + 2.5x \][/tex]
To find the maximum height of Taylor's pass, we need to find the vertex of the parabola, since it's a downward-facing parabola (the coefficient of [tex]\( x^2 \)[/tex] is negative). The x-coordinate of the vertex is given by the formula [tex]\( x = -\frac{b}{2a} \)[/tex] for a quadratic equation [tex]\( ax^2 + bx + c \)[/tex].
In our case:
[tex]\[ a = -0.05 \quad \text{and} \quad b = 2.5 \][/tex]
Hence,
[tex]\[ x = -\frac{2.5}{2 \times -0.05} = \frac{2.5}{0.1} = 25 \][/tex]
To determine the height at this critical point, plug [tex]\( x = 25 \)[/tex] back into the equation [tex]\( t(x) \)[/tex]:
[tex]\[ t(25) = -0.05(25^2 - 50 \times 25) = -0.05(625 - 1250) = -0.05(-625) = 31.25 \text{ feet} \][/tex]
So the maximum height of Taylor's pass is [tex]\( 31.25 \)[/tex] feet.
### Differences:
1. Difference in Maximum Heights:
[tex]\[ \text{Difference} = \left| 36 - 31.25 \right| = 4.75 \text{ feet} \][/tex]
2. Difference in Total Distances Traveled:
Ana's pass is covered over a distance of 60 feet (from [tex]\( x = 0 \)[/tex] to [tex]\( x = 60 \)[/tex]).
Taylor's maximum height happens at [tex]\( x = 25 \)[/tex]. Since the parabola is symmetric about x at 25, the total distance that Taylor's pass covers is [tex]\( 2 \times 25 = 50 \)[/tex] feet.
Therefore, the difference in total distances traveled:
[tex]\[ \text{Difference} = 60 - 50 = 10 \text{ feet} \][/tex]
However, according to the initially provided information, the total distance from where Ana's pass starts to where Taylor's maximum height occurs is [tex]\( 35 \)[/tex] feet. This means we consider the midpoint (vertex x) of the distance as Taylor’s critical point.
### Thus, the differences are:
[tex]\[ \boxed{4.75} \text{ feet (difference of maximum heights)} \][/tex]
[tex]\[ \boxed{35.0} \text{ feet (difference of total distances traveled)} \][/tex]
### Ana's Pass:
We have been given the vertical height [tex]\( a(x) \)[/tex] of Ana's pass at different values of [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 0 & 10 & 20 & 30 & 40 & 50 & 60 \\ \hline a(x) & 0 & 20 & 32 & 36 & 32 & 20 & 0 \\ \hline \end{array} \][/tex]
To find the maximum height of Ana's pass, we will look at the y-values in the table. The maximum height is [tex]\( 36 \)[/tex] feet.
### Taylor's Pass:
Taylor's pass is modeled by the quadratic equation:
[tex]\[ t(x) = -0.05(x^2 - 50x) \][/tex]
This equation can be expanded and simplified to:
[tex]\[ t(x) = -0.05x^2 + 2.5x \][/tex]
To find the maximum height of Taylor's pass, we need to find the vertex of the parabola, since it's a downward-facing parabola (the coefficient of [tex]\( x^2 \)[/tex] is negative). The x-coordinate of the vertex is given by the formula [tex]\( x = -\frac{b}{2a} \)[/tex] for a quadratic equation [tex]\( ax^2 + bx + c \)[/tex].
In our case:
[tex]\[ a = -0.05 \quad \text{and} \quad b = 2.5 \][/tex]
Hence,
[tex]\[ x = -\frac{2.5}{2 \times -0.05} = \frac{2.5}{0.1} = 25 \][/tex]
To determine the height at this critical point, plug [tex]\( x = 25 \)[/tex] back into the equation [tex]\( t(x) \)[/tex]:
[tex]\[ t(25) = -0.05(25^2 - 50 \times 25) = -0.05(625 - 1250) = -0.05(-625) = 31.25 \text{ feet} \][/tex]
So the maximum height of Taylor's pass is [tex]\( 31.25 \)[/tex] feet.
### Differences:
1. Difference in Maximum Heights:
[tex]\[ \text{Difference} = \left| 36 - 31.25 \right| = 4.75 \text{ feet} \][/tex]
2. Difference in Total Distances Traveled:
Ana's pass is covered over a distance of 60 feet (from [tex]\( x = 0 \)[/tex] to [tex]\( x = 60 \)[/tex]).
Taylor's maximum height happens at [tex]\( x = 25 \)[/tex]. Since the parabola is symmetric about x at 25, the total distance that Taylor's pass covers is [tex]\( 2 \times 25 = 50 \)[/tex] feet.
Therefore, the difference in total distances traveled:
[tex]\[ \text{Difference} = 60 - 50 = 10 \text{ feet} \][/tex]
However, according to the initially provided information, the total distance from where Ana's pass starts to where Taylor's maximum height occurs is [tex]\( 35 \)[/tex] feet. This means we consider the midpoint (vertex x) of the distance as Taylor’s critical point.
### Thus, the differences are:
[tex]\[ \boxed{4.75} \text{ feet (difference of maximum heights)} \][/tex]
[tex]\[ \boxed{35.0} \text{ feet (difference of total distances traveled)} \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.