LINR 1 | Lesson 2 | Practice (Equations of Linear Situations) Solution

A Linear Situation:

  1. \(y=4ox+4500\); where \(y=\)elevation and \(x=\)number of hours.
  2. \(y = 40(8)+4500\); 4820 ft
  3. \(5020 = 40x + 4500\); 13 hours

Table:

  1. \(y=67 – 4x\); where \(x=\)hours and \(y =\)degrees
  2. \(y=67 – 4(2.5)\); 57 degrees
  3. \(37 = 67 -4x\); 7.5 hrs

Graph:

  1. \(y = 40 + \Large\frac{x}{4}\); \(x =\)number of chirps per min and \(y=\) temperature
  2. \(y = 40 + \Large\frac{300}{4}\); 115 degrees
  3. \(62 = 40+\Large\frac{x}{4}\); 88 chirps

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