Question

robin is a competitive archer. the graph below is a record of robins last three arrows, where (0, 0) is the center of the bullseye. graph units are in centimeters. what is the distance of robins closest arrow from the center of the bullseye? round the final answer to the nearest tenth of a centimeter. do not round intermediate calculations.

Explanation:

Step1: Identify coordinates

Arrow A is at (-4, 4), Arrow B is at (4, - 4), Arrow C is at (-2,-6). The center of the bullseye is at (0,0).

Step2: Use distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For a point $(x,y)$ and the origin $(0,0)$, the formula simplifies to $d=\sqrt{x^{2}+y^{2}}$.

Step3: Calculate distance for Arrow A

For Arrow A with $(x=-4,y = 4)$, $d_A=\sqrt{(-4)^{2}+4^{2}}=\sqrt{16 + 16}=\sqrt{32}$.

Step4: Calculate distance for Arrow B

For Arrow B with $(x = 4,y=-4)$, $d_B=\sqrt{4^{2}+(-4)^{2}}=\sqrt{16 + 16}=\sqrt{32}$.

Step5: Calculate distance for Arrow C

For Arrow C with $(x=-2,y=-6)$, $d_C=\sqrt{(-2)^{2}+(-6)^{2}}=\sqrt{4 + 36}=\sqrt{40}$.

Step6: Compare distances

Since $\sqrt{32}<\sqrt{40}$, we focus on $\sqrt{32}$. Simplify $\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\approx5.7$ (rounded to the nearest tenth).

Answer:

$5.7$