An archer

An archer stands 60 meters (m) from a target. She launches an arrow that lands 3 centimeters (cm) from the bull's eye. The archer changes her position to 40 m from the target, and her next arrow lands 2 cm from the bull's eye. She changes her position to 20 m, and her next arrow lands 1 cm from the bull's eye.

Which points describe the situation when the archer stands at 40 m?

Let the x-coordinate be the archer's distance from the target in meters, and let the y-coordinate be the arrow's distance from the bull's eye in centimeters.

Final Answer:

e : e=x^2/600-x/2+4/3=y

Step-by-step explanation:

$$\begin{gathered} e(50)=3 \\ e(40)=2 \\ e(20)=1 \\ y=e(x)=a_x^2+b_x+c \\ 3=a\cdot 50^2+b\cdot 50+c \\ 2=a\cdot 40^2+b\cdot 40+c \\ 1=a\cdot 20^2+b\cdot 20+c \\ 2500a+50b+c=3 \\ 1600a+40b+c=2 \\ 400a+20b+c=1 \\ Row2-\frac{1600}{2500}\cdot Row1\to Row2 \\ 8b+0.36c=0.08 \\ Row3-\frac{400}{2500}\cdot Row1\to Row3 \\ 12b+0.84c=0.52 \\ Pivot:Row2\leftrightarrow Row3 \\ Row3-\frac{8}{12}\cdot Row2\to Row3 \\ -0.2c=-0.267 \\ c=\frac{-0.26666667}{-0.2}=1.33333333 \\ b=\frac{0.52-0.84c}{12}=\frac{0.52-0.84\cdot 1.33333333}{12}=-0.05 \\ a=\frac{3-50b-c}{2500}=\frac{3-50\cdot (-0.05)-1.33333333}{2500}=0.00166667 \\ a=\frac{1}{600}\doteq 0.001667 \\ b=\frac{-1}{20}=-0.05 \\ c=\frac{4}{3}\doteq 1.333333 \\ e:e=x^2/600-x/2+4/3=y \end{gathered}$$

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