Question

a taxi ride costs p40.00 for the first 500 meters , and each additional 300 meters (or a fraction thereof )adds p3.50.to the fare .use a peicewise function to represent the taxi fare in terms of the distance d in meters

Answer 1

Answer:

Function representing taxi fair , $f(d)=3.50\times(\frac{d-500}{300})+40$   d > 500.

Step-by-step explanation:

Given:

Rs. 40.00 for first 500 m

Rs. 3.50 for each additional 300 m.

we need to write the equation representing taxi fair in term os distance.

Let the distance of the ride = d meters.

So,

$taxi\:fare=3.50\times(\frac{d-500}{300})+40$      d > 500

Therefore, Function representing taxi fair , $f(d)=3.50\times(\frac{d-500}{300})+40$   d > 500.

Answer 2

Answer:

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Step-by-step explanation:

Function representing taxi fair , f(d)=3.50\times(\frac{d-500}{300})+40f(d)=3.50×(

300

d−500

)+40 d > 500.

Step-by-step explanation:

Given:

Rs. 40.00 for first 500 m

Rs. 3.50 for each additional 300 m.

we need to write the equation representing taxi fair in term os distance.

Let the distance of the ride = d meters.

So,

taxi\:fare=3.50\times(\frac{d-500}{300})+40taxifare=3.50×(

300

d−500

)+40 d > 500

Therefore, Function representing taxi fair , f(d)=3.50\times(\frac{d-500}{300})+40f(d)=3.50×(

300

d−500

)+40 d > 500.