Respuesta :
Answer:
The correct option is d.
Step-by-step explanation:
The approximate population P, in thousands, for a species of frogs in a particular rain forest, x years after 1999 is given by the formula
[tex]P=0.672x^2-0.046x+3[/tex]
We need to find the year it which the population reach 182 frogs.
Substitute P=182 in the given formula.
[tex]182=0.672x^2-0.046x+3[/tex]
Subtract 182 from both the sides.
[tex]0=0.672x^2-0.046x+3-182[/tex]
[tex]0=0.672x^2-0.046x-179[/tex]
Multiply both sides by 1000 to remove decimals.
[tex]0=672x^2-46x-179000[/tex]
Quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substitute a=672, b=-46 and c=-179000 in the quadratic formula.
[tex]x=\frac{-\left(-46\right)\pm\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}[/tex]
On simplification we get
[tex]x=\frac{-\left(-46\right)+\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx 16.355[/tex]
[tex]x=\frac{-\left(-46\right)-\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx -16.287[/tex]
The value of x can not be negative because x is number of years after 1999.
x=16.35 in means is 17th year after 1999 the population reach 182 frogs.
[tex]1999+17=2016[/tex]
The population reach 182 frogs in 2016. Therefore the correct option is d.
