What is the approximate length of the hypotenuse of a right triangle with vertices (-6, 2),
(6,-3) and (-6, -3)?

Step-by-step explanation:

A(-6, 2),   B(6,-3) and C(-6, -3)

AB² =  (x₂ - x₁)² + (y₂ - y₁)²

      = ( 6 -[-6])² + ( -3 -2)²

      = ( 6 + 6)² + ( -3 -2)² = 12²  + (-5)²  = 144 + 25 =169

AB = √169 = 13 units

BC²  = ( -6 -6)²  + ( -3 - [-3])² = (-6-6)²  + (-3 +3)²

      = (-12)²  + 0  = 144

BC = √144 = 12 unis

CA²  = (-6 - [-6])²  +(-3-2)²  = (-6 + 6)²  + ( -3-2)²

         = 0 + (-5)²  = 25

CA =√25 = 5 units

length of the hypotenuse of a right triangle  = 13units

the answer is 4.8 units on e2020.

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).