Alake has a native population of 1,000 frogs. the native frog population increases at a rate of 200 frogs per year. a new species of 10 frogs migrates to the lake. this new species’ population increases at a rate of 50% per year. when will the two populations equal each other? select to reveal each frog population’s growth model.

a system of equations can be created with the two functions to determine when the populations will have the same population output value, y.
y = 200x + 1,000
y = 10(1.5x)

to determine when the populations will be equal, set the equations equal to each other, and solve for x.
200x + 1,000 = 10(1.5x)

this equation is a little more challenging to solve by algebraic methods. in this lesson, you will solve systems of equations like the one above approximately by effectively using technology and tables.

1. Alright lets start off with the key points for example:

-native population of frogs is 1000

-200 new frogs per year of native frogs

-10 new frogs migrate to the lake

-50% more new frogs of the current population

So what I did was make a graph of the 2 equations for example:

y = 200x + 1000

y = 1.5^x · 10

Then you put them together to make 1 equation which is:

200x + 1000 = 1.5^x · 10

Which gives us the answer of 14.745 or rounded up to 15.

d. millimeters per hour.

step-by-step explanation:

we are given that,

carlo used a rain gauge to measure the amount of water his lawn is receiving from the sprinklers.

since, according to the options, we see that,

water cannot be measured by 'miles per hour', 'meters per second' and 'grams per seconds'.

we get that, the water can be measured by 'millimeters per hour'.

hence, the appropriate rate for describing the amount of water is 'millimeters per hour'.