## Algebra 1

a) Please see the graph. b) 10 miles: $s=100$, 1 mile: $s=1600$, .1 mile: $s=16000$ c) The closer a radio is to a transmitter, the stronger the frequency. Since you are closer to one transmitter than to the other, then the weaker frequency would be muffled by the stronger frequency.
a) $d \geq 40$ when $s \leq 1$ b) $s=1600/d^2$ $s = 1600/10^2$ $s=1600/100$ $s=160$ $s=1600/d^2$ $s=1600/1^2$ $s=1600/1$ $s=1600$ $s=1600/d^2$ $s=1600/.1^2$ $s=1600/.01$ $s=16000$