Question

1. a biologist recorded the population growth of two species of frogs in a pond area starting in 2010. the first species started with 42 frogs and increased by 6 frogs per year. the second species started with 30 frogs and increased by 8 frogs per year.

a. write an equation modeling equal populations of the frogs t years after 2010.
b. in what year were the populations of the two groups of frogs equal?
use the information to answer problems 6 - 7.
riley is comparing cell phone plans. the table shows four options riley is considering. the gigabytes of data used each month is represented by g.

plan	monthly fee	charge per gb	total monthly cost ($)
2	$40	$10	40 + 10g
3	$80	$0	80
4	$50	$5	50 + 5g

Explanation:

Step1: Find population - growth equations for each species

For the first species, the initial population $a_1 = 42$ and the rate of increase $r_1=6$ per year. So the population $P_1$ after $t$ years is $P_1 = 42 + 6t$. For the second species, the initial population $a_2 = 30$ and the rate of increase $r_2 = 8$ per year. So the population $P_2$ after $t$ years is $P_2=30 + 8t$. When the populations are equal, $P_1 = P_2$. So the equation is $42+6t=30 + 8t$.

Step2: Solve the equation for $t$

Subtract $6t$ from both sides: $42=30 + 8t-6t$, which simplifies to $42=30 + 2t$. Then subtract 30 from both sides: $42 - 30=2t$, so $12 = 2t$. Divide both sides by 2: $t=\frac{12}{2}=6$.

Step3: Find the year

Since $t$ is the number of years after 2010, the year is $2010 + 6=2016$.

Answer:

A. $42 + 6t=30 + 8t$
B. 2016