In a large city, taxicabs charge $1.00 for the first 1÷5 mile and $0.36 for each additional 1÷5 mile. frank has only 9.50. what is the maximum distance he can travel (not including a tip for the cabbie)?

Frank can travel 4.8 mi.

The cabbie measures the distance in fifths of a mile.  Let’s say that Frank travels n fifths.

The charge(C) is $1.00 for the first fifth and $0.36 for the remaining (n – 1) fifths.

C = 1.00 + 0.36(n-1)

C = 1.00 + 0.36n -0.36 = 0.64 + 0.36n.

If Frank can spend only $9.50,

9.50 = 0.64 + 0.36n

0.36n = 9.50 – 0.64 = 8.86

n = 8.86/0.36 = 24.6

If the meter “pulses” after each fifth, Frank can travel only an integral number of fifths (24).

Frank can travel 24 × ¹/₅ mi = 4.8 mi

Check: $1 + $0.36×23 = $9.28. Frank will have $0.22 left over, not enough for another fifth-mile.

3x + 18 = 147°

step-by-step explanation:

3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.

i hope i you.