The center of an ice rink is located at $$ (0,0) $$ on a coordinate system measured in feet. Sandi skates along a path that can be modeled by the equation $$ y=0.08 x^{2}-1.6 x+13 $$. David starts at $$ (20,13) $$ and skates along a path that can be modeled by a quadratic function with a vertex at $$ (10,5) $$. Which system of equations models the paths of the two skaters? $$ y=0.08 x^{2}-1.6 x+13 \text { and } y=0.08(x-10)^{2}+5 $$ $$ y=0.08 x^{2}-1.6 x+13 $$ and $$ y=1.6(x-10)^{2}+5 $$ $$ y=0.08 x^{2}-1.6 x+13 $$ and $$ y=0.8(x-10)^{2}+5 $$ $$ v=0.08 x^{2}-1.6 x+13 $$ and $$ v=5(x-10)^{2}+5 $$

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UpStudy AI · Accepted Answer

We are given Sandi's path by the equation  
$$
y = 0.08x^2 - 1.6x + 13.
$$

David's path is described by a quadratic function with a vertex at $$(10,5)$$. In vertex form, this function can be written as  
$$
y = a(x-10)^2 + 5.
$$

David starts at $$(20,13)$$, so his equation must satisfy  
$$
13 = a(20-10)^2 + 5.
$$

Step 1. Compute the squared term:  
$$
(20-10)^2 = 10^2 = 100.
$$

Step 2. Substitute and solve for $$a$$:  
$$
13 = 100a + 5.
$$

Step 3. Isolate $$a$$:  
$$
100a = 13 - 5 = 8,
$$
$$
a = \frac{8}{100} = 0.08.
$$

Step 4. Write David's equation with the determined $$a$$:  
$$
y = 0.08(x-10)^2 + 5.
$$

Thus, the system of equations modeling the paths of the two skaters is:
$$
\begin{cases}
y = 0.08x^2 - 1.6x + 13,\[1mm]
y = 0.08(x-10)^2 + 5.
\end{cases}
$$
The system of equations modeling the paths of Sandi and David is: $$ \begin{cases} y = 0.08x^2 - 1.6x + 13,\ y = 0.08(x-10)^2 + 5. \end{cases} $$

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