Researchers have concluded that a dry basin began to fill with water in 1880. the water level rose an average of 1.6 millimeters (mm) per year from 1880 to 2009. the rise in water level since 1880 can be modeled by f(x) = 1.6x, where x is the number of years since 1880 and f is the total rise of the water level in mm.

1. what is the domain of f(x)?
2. by how much did the water level rise in all from 1880 to 1950?
3. in what year was the water level 100 mm higher than in 1880?

remember: 1 m = 1,000 mm

and

Step-by-step explanation:

1. The function is good for years 1880 to 2009, 0 to 129 years after 1880. Values of x can be anything in the domain [0, 129].

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2. 1950 -1880 = 70. The year 1950 corresponds to x=70, so the function tells us the water level rose ...

  f(70) = 1.6·70 = 112 . . . . . mm

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3. We want to find x when f(x) = 100. That will be the solution to ...

  100 = 1.6x

  100/1.6 = x = 62.5

Then 62.5 years after 1880 is year 1942.5. The water level was 100 mm higher than in 1880 in the year 1942.

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.  

the answer is d on edge

step-by-step explanation: