[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{5}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{18}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(18-5)^2+(3-14)^2}\implies d=\sqrt{13^2+(-11)^2} \\\\\\ d=\sqrt{169+121}\implies d=\sqrt{290}\implies d\approx 17.03[/tex]
Answer:
About 17
Step-by-step explanation:
First, find the distance between the x coordinates (13) and the distance between the y coordinates (11). Next, plug these numbers into the Pythagorean theorem (a^2+b^2=c^2) and solve. 13 squared is 169 and 11 squared is 121. Add those together to get 290. The square root of 290 is about 17.02938637 (that's how many digits my calculator can show), and it has no simplified radical form.
