If the radius of a cylinder was shrunk down to a quarter of its orignal size and the height was reduced to a third of its original size, what would be the formula to find the modified surface area?

The modified area is (1/48) (2πr(4h+3r))

Step-by-step explanation:

Let the radius be 'r' and height be 'h'.

Area of cylinder= 2π r(h+r)

The radius is shrunk down to quarter of its original radius

 r = r/4

The height is reduced to a third of its original height

h = h/3

New Area = 2π(r/4) [(h/3) +(r/4) ]

= (1/4)2πr[(4h+3r) /12]

= (1/48) (2πr(4h+3r))

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a