Question

me ayudan con trigonometriaa??

Answer 1

Respuesta:

me ayudan con trigonometriaa??

Explicación paso a paso:

$(3,-4);(6,1)\\d=\sqrt{\left(3-6\right)^2+\left(-4-1\right)^2}\\d=\sqrt{3^2+5^2}\\d=\sqrt{9+5^2}\\d=\sqrt{34}\\=5.83$

$(8,0);(4,3)\\d=\sqrt{\left(8-4\right)^2+\left(0-3\right)^2}\\d=\sqrt{4^2+3^2}\\d=\sqrt{16+3^2}\\d=\sqrt{16+9}\\d=\sqrt{5^2}\\d=5$

$(6,4);(0,-1)\\d=\sqrt{\left(6-0\right)^2+\left(4-\left(-1\right)\right)^2}\\d=\sqrt{\left(6-0\right)^2+\left(4+1\right)^2}\\d=\sqrt{6^2+\left(4+1\right)^2}\\d=\sqrt{6^2+5^2}\\d=\sqrt{36+5^2}\\d=\sqrt{36+25}\\d=\sqrt{61}\\=7.81$