Records indicate that x years after 2008, the average property tax on a three bedroom home in a certain community was T(x) =20x^2+40x+600 dollars. Required:
a. At what rate was the property tax increasing with respect to time in 2008?
b. By how much did the tax change between the years 2008 and 2012?

a) 40 dollars

b) 480 dollars

Step-by-step explanation:

Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0

T'(x) = 2(20)x¹ + 40x° + 0

T'(x) = 40x+40

At x = 0,

T'(0) = 40(0)+40

T'(0) = 40

Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).

b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)

Given T(x) =20x²+40x+600

T(4) =20(4)²+40(4)+600

T(4) = 320+160+600

T(4) = 1080 dollars

Also T(0) =20(0)²+40(0)+600

T(0) = 0+0+600

T(0)= 600 dollars

T(4) - T(0) = 1080 - 600

T(4) - T(0) = 480 dollars

Hence, the tax has changed by $480 between 2008 and 2012

the lengths of the bases of the trapezoids are
trapezoid A
b=19cm, and B=35cm, b the short base and B the long base

trapezoid B
b=19cm, and B=35cm, b the short base and B the long base

the height is h=15cm

the surface of A is A= hx(b+B) /2

so the surface is A= 15(19+35)/2=405cm²
we know that suface A =surfaceB, so Surface B= 405cm²

finally the total surface painted is 2 x 405cm²= 810cm²

the perimeter of the trapezoid A is PA=17+19+35+17=88cm
the perimeter of the trapezoid B is PB=17+19+35+17=88cm

 one piece of sandpaper will not be enough to sand the edges of one trapezoid