A balanced diet is one that includes a variety of foods in the right amounts and ratios to meet the body's needs for calories, proteins, minerals, vitamins, and other nutrients. A small portion of the diet is also set aside for extra nutrients to last through the brief period of leanness. In this situation, we must choose a unique menu that offers the

In this context, we have to first prepare the list of the number of carbohydrates, lipids, and proteins and then we have to identify the maximum amount required in the meal. then observe the common food in the school.

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In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,2

Alisiya [41]

Answer:

Suppose that we have a given function f(x)

The average rate of change of the function between two values x₁ and x₂ is given by:

$r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

a) We want to find the average (rate) of change on the size of population from 1985 to 1995.

We have that:

f(1985) = 750

f(1995) = 1500

Then we have:

$r = \frac{1500 - 750}{1995 - 1985} = 750/10 = 75$

This means that the population of trees increases, in average, at a rate of 75 trees per year.

b) What is the density of trees each year that they were counted?

This will be equal to the quotient between the number of trees and the area.

1985: number of trees = 750 pines

area = 250 ha

Then the density is:

D(1985) = (750 pines)/(250 ha) = 3 pines/ha

So 1985, there were 3 pines per hectare.

1990: number of trees = 1250 pines

area = 250 ha

Then the density is:

D(1990) = (1250 pines)/(250 ha) = 5 pines/ha

1995: number of trees = 1500 pines

area = 250 ha

The density is:

D(1995) = (1500 pines)/(250 ha) = 6 pines/ha

3) now we want to get the average change between 1985 and 1995 in the density, this will be:

$r = \frac{D(1995) - D(1885)}{1995 - 1985} = \frac{6 pines/ha - 3pines/ha}{10} = 0.3 pines/ha$

So, on average, each year the number of pines per hectare increases by 0.3

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3 years ago