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.

25 questions , because they start with 100.  

350-100=250 .  

. 250/10=25

step-by-step explanation:

first take off 100, then what x 10 = 250? that is your answer. (25)

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: