The number of E.coli bacteria cells in a pond of stagnant water can be represented by the function below, where A represents the
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Let Daniel's age be 'x' years , hence, Lisa's age is (x - 5) years. Let Grandma's age be 'y' = 54 years.
Hence,
x(x - 5) = 2/3 × y
x² - 5x = 2 × 54 years / 3
x² - 5x = 36 years
x² - 5x - 36 = 0
∆ = (-5)² - 4(1)(-36) = 25 + 144 = 169
√∆ = √169 = 13
x1 = (5+13)/2, x2 = (5-13)/2
x1 = 9, x2 = -4
Since, age can't be negative,
Daniel's age = x = 9 years.
Hence,
x(x - 5) = 2/3 × y
x² - 5x = 2 × 54 years / 3
x² - 5x = 36 years
x² - 5x - 36 = 0
∆ = (-5)² - 4(1)(-36) = 25 + 144 = 169
√∆ = √169 = 13
x1 = (5+13)/2, x2 = (5-13)/2
x1 = 9, x2 = -4
Since, age can't be negative,
Daniel's age = x = 9 years.