Maintenance costs for an apartment building generally increase as the building ages. From past records, it is estimated that the rate of increase in maintenance costs for this type of building is approximated by Where x is the age of the complex in years and M (x) is the total accumulated cost of maintenance for x years. Write a definite integral that will give the total maintenance costs from the end of the second year to the end of the seventh year of use and evaluate it.

M(x) = 29480 units

Step-by-step explanation:INCOMPLETE QUESTION.

Information describing the equation of maintenance costs for the type of building is needed, we can either work with a maintenance costs equation

or with any other equation from which to get the maintenance costs equation.

Particularly similar statement problem give M´(x) = 90*x² + 5000, therefore if we integrate such equation we get M(x) as maintenance cost equation. We will do that then

M´(x)  = 90*x² +  5000

∫M´(x)  =  M(x)  = ∫(90*x² +  5000 ) dx

but we require costs from the end of second year up to the end of the 7 th year then

M(x)  = ∫₃⁷ (90*x²  +  5000) dx

M(x)  = (90/3 )*x³   +  5000*x) |₃⁷

M(x)  = 30*(7)³ + 5000* (7)  - 30*(3)³ - 5000*3

M(x)  =  10290 + 35000 - 810 - 15000

M(x)  =  29480 $

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.