Amanufacturer wants to double the volume of a 3 in.×2 in.×6 in. 3 i n . × 2 i n . × 6 i n . box, while using as little extra cardboard as possible. which statement is true?

2 inch dimension will give smallest increase.

Step-by-step explanation:

Length = 3 in

width = 2 in

height = 6 in

Extra cardboard  means to find surface area

on doubling the length

length = 6 In

width = 2 In

Height = 6In

Surface area for the above dimensions = 2 [ 6x2+2x6+6x6] = 120 sq in

On doubling the width

length = 3 in

width = 4 in

Height = 6 inch

Surface area for the above dimensions= 2 [ 3x4+4x6+6x3] = 2[54] = 108 sq inches

On doubling height

Length =3 in

width = 2 in

Height = 12 in

Surface area for above dimensions = 2 [ 3x2+2x12+12x3] = 2[6+24+36] = 132 sq inch

On doubling width surface area is minimum.

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.  

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !