$3^{-3}= \frac{1}{3^3}= \frac{1}{27} \\ \\ 3^{-2}= \frac{1}{3^2}= \frac{1}{9} \\ \\ 3^{-1}= \frac{1}{3^1}= \frac{1}{3} \\ \\3^{0}=1 \\\\ 3^1=3\\\\3^2=9\\\\3^3=27$

7 0

3 years ago

An engineer is surveying a building with a device that is 5 feet above the ground. He is 50 feet from the building and is lookin

dolphi86 [110]

1. Check the sketch of the problem.

2. In right triangle DCE,

$tan50= \frac{EC}{DC}= \frac{x}{50}$

tan 50 can be found to be 1.19 using a scientific calculator or other software.

$1.19=\frac{x}{50}$

x=1.19*50=59.5 (feet)

3. The height of the building is 59.5+5=64.5 (feet)

Answer: 64.5 (feet)

5 0

2 years ago