Lyra and Donna are testing the two-way radios they built for their high school science project. Lyra goes to the top of a building that is 22 meters high with one of the radios and stands directly above the building’s entrance. Donna walks out of the entrance, perpendicular to the building, with the other radio. If Donna is more than 50 meters from the entrance, the connection between the radios breaks. What is the range of communication for the two radios to the nearest meter? PS follow me on only fans angela_esparza9

Lyra is 22 meters above the entrance.

The radios work until Donna is 50 meters away from the entrance.

At that moment, Lyra, Donna, and the entrance form a right triangle.

The hypotenuse is the actual distance between Lyra and Donna.

          Distance  =  √ (22² + 50²)

                        =   √ (484 m² + 2,500 m²)

                        =    √ (2984 m²)

                        =        54.63 meters.

Rounded to the nearest meter,    Distance = 55 meters.

THAT's the maximum point-to-point range of the radios.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3