A company is replacing cables with fiber optic lines in rectangular casing BCDE. If segment DE = 3 cm and segment BE = 3.5 cm, what is the smallest diameter of pipe that will fit the fiber optic line? Round your answer to the nearest hundredth.

3.66 cm

Step-by-step explanation: Given a rectangular casing BCDE with segment DE = 3 cm and segment BE = 3.5 cm.

The area A of a rectangle is length multiply by width.

Where length L = 3.5 cm and

width W = 3 cm

Area A = 3.5 × 3 = 10.5 cm^2

The pipe that will fit the fiber optic line is in cylindrical shape. Where area of a cylinder = πr^2.

But area A = 10.5. Substitute the values for the area of the cylinder

10.5 = πr^2

10.5 = 3.143 × r^2

Make r^2 the subject of formula

r^2 = 10.5/3.143

r = sqrt ( 3.34225 )

r = 1.828

Diameter = 2 × radius

Diameter = 2 × 1.829

Diameter = 3.656 cm

Therefore, the smallest diameter of pipe that will fit the fiber optic line is 3.66 cm approximately.

bull

step-by-step explanation:

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.