Question

mia is a competitive dart player. the graph below is a record of mias last three darts, where (0, 0) is the center of the bullseye. graph units are in centimeters. what is the distance of mias farthest dart 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 dart coordinates

Dart A: $(- 2,-6)$, Dart B: $(-7,0)$, Dart C: $(4,-5)$

Step2: Use distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ with $(x_1,y_1)=(0,0)$

For Dart A: $d_A=\sqrt{(-2 - 0)^2+(-6 - 0)^2}=\sqrt{4 + 36}=\sqrt{40}$
For Dart B: $d_B=\sqrt{(-7 - 0)^2+(0 - 0)^2}=\sqrt{49}=7$
For Dart C: $d_C=\sqrt{(4 - 0)^2+(-5 - 0)^2}=\sqrt{16 + 25}=\sqrt{41}$

Step3: Compare distances

Since $\sqrt{40}\approx6.3$, $\sqrt{41}\approx6.4$, and $7>6.4>6.3$

Answer:

$7.0$