5. a biologist recorded the population growth of two species of frogs in a pond area starting in 2010. the…

5. 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.\na. write an equation modeling equal populations of the frogs t years after 2010.\nb. in what year were the populations of the two groups of frogs equal?\nuse the information to answer problems 6 - 7.\nriley is comparing cell phone plans. the table shows four options riley is considering. the gigabytes of data used each month is represented by g.\n| plan | monthly fee | charge per gb | total monthly cost ($) |\n| ---- | ---- | ---- | ---- |\n| 1 | $60 | $5 | 60 + 5g |\n| 2 | $40 | $10 | 40 + 10g |\n| 3 | $80 | $0 | 80 |\n| 4 | $50 | $5 | 50 + 5g |

5. 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.\na. write an equation modeling equal populations of the frogs t years after 2010.\nb. in what year were the populations of the two groups of frogs equal?\nuse the information to answer problems 6 - 7.\nriley is comparing cell phone plans. the table shows four options riley is considering. the gigabytes of data used each month is represented by g.\n| plan | monthly fee | charge per gb | total monthly cost ($) |\n| ---- | ---- | ---- | ---- |\n| 1 | $60 | $5 | 60 + 5g |\n| 2 | $40 | $10 | 40 + 10g |\n| 3 | $80 | $0 | 80 |\n| 4 | $50 | $5 | 50 + 5g |

Answer

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