What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?

Answer:25.6 unitsStep-by-step explanation: From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}where(x_1,y_1) are the coordinates of the first point(x_2,y_2) are the coordinates of the second point- For AB:d=\sqrt{[1-(-5)]^{2}+(4-4)^2}d=\sqrt{(1+5)^{2}+(0)^2}d=\sqrt{(6)^{2}}d=6- For BC:d=\sqrt{(3-1)^{2} +(-4-4)^{2}}d=\sqrt{(2)^{2} +(-8)^{2}}d=\sqrt{4+64}d=\sqrt{68}d=8.24- For AC:d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}d=\sqrt{(3+5)^{2} +(-8)^{2}}d=\sqrt{(8)^{2} +64}d=\sqrt{64+64}d=\sqrt{128}d=11.31Next, now that we have our lengths, we can add them to find the perimeter of our triangle:p=AB+BC+ACp=6+8.24+11.31p=25.55p=25.6We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.Read more on Brainly.com -