Oil spilled from a ruptured tanker spreads in a circle. (Three separate, yet related, problems.) (i) If the area of the circle increases at a constant rate of 4 miles squared per hour, how fast is the radius of the spill increasing when the area is 15 miles squared

0.29 mi / hr

Step-by-step explanation:

Given that:

Area (A) = 15mi²

Area increases at a constant rate of 4mi²/ hr

Area(A) of a circle is given as πr² - - - (1)

Derivative of A with respect to time ;

A = πr²

dA/dt = 2πr(dr/dt) - - - (2)

Rate of change of Area with time (dA/dt) = 4mi²/hr

From (1) :

A = πr²

A = 15mi²

15 = πr²

15/π = r²

r = √(15/π)

From (2) :

dA/dt = 2πr(dr/dt)

4 = 2π√(15/π)dr/dt

dr/dt = 4/2π√(15/π)

= 0.29134

= 0.29 mi / hr

answer:     true

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet