Question

16. FINANCIAL LITERACY A specific car's fuel economy in miles per gallon can be

approximated by f(x) = 0.00000056x4 - 0.000018x3 - 0.016x2 + 1.38x - 0.38,

where x represents the car's speed in miles per hour. Determine the fuel economy

when the car is traveling 40, 50, and 60 miles per hour.

Answer 1

To determine the fuel economy of a car at 40, 50, and 60 mph, substitute the speeds into the given polynomial function and calculate the resulting values to get the car's fuel efficiency in miles per gallon for each speed.

Step-by-step explanation:

The subject of the student's question involves calculating the fuel economy of a car at different speeds using a given polynomial function. To determine the fuel economy when the car is traveling at 40, 50, and 60 miles per hour, the speeds must be substituted into the fuel economy function f(x) and the resulting value will represent the car's fuel efficiency at that speed in miles per gallon (mpg).

To calculate:

1. For x=40: f(40) = 0.00000056(40)^4 - 0.000018(40)^3 - 0.016(40)^2 + 1.38(40) - 0.38
2. For x=50: f(50) = 0.00000056(50)^4 - 0.000018(50)^3 - 0.016(50)^2 + 1.38(50) - 0.38
3. For x=60: f(60) = 0.00000056(60)^4 - 0.000018(60)^3 - 0.016(60)^2 + 1.38(60) - 0.38

After performing the calculations for each speed, the resulting values will give the approximate fuel economy of the car at 40, 50, and 60 mph, respectively.

Answer 2

a) 29.50 miles per gallon b) 29.87 miles per gallon c) 28.19 miles per gallon

Step-by-step explanation:

A specific car fuel economy (miles/gallon) = f(x) = 0.00000056x^4 - 0.000018x^3 - 0.016x^2 + 1.38x - 0.38

where x represents the car speed in miles per hour.

when x = 40 miles per hour

A specific car fuel economy (miles/gallon) = f(x) = 0.00000056(40)^4 - 0.000018(40)^3 - 0.016(40)^2 + 1.38(40) - 0.38 = 1.4336 - 1.152 - 25.6 + 55.2 - 0.38 = 29.50 miles per gallon

Similarly,

when x = 50

A specific car fuel economy (miles/gallon) = f(x) = 0.00000056(50)^4 - 0.000018(50)^3 - 0.016(50)^2 + 1.38(50) - 0.38 = 3.5 - 2.25 - 40 + 69 - 0.38 = 29.87 miles per gallon

Similarly,

when x = 60

A specific car fuel economy (miles/gallon) = f(x) = 0.00000056(60)^4 - 0.000018(60)^3 - 0.016(60)^2 + 1.38(60) - 0.38 = 7.26 - 3.89 - 57.6 + 82.8 - 0.38 = 28.19 miles per gallon