A music company wants to design a cardboard box for mailing a 22-CD set measuring 14.2 cm by 12.4 cm by 2.5 cm. what amount of suface area will the box have? ​

At first students may find the sum of all the areas of the faces.

$$ \begin{alignat}{3} \text{Front: } (5)(9.5) &=47.5 \text{ in}^2 &\qquad \text{Top: } (5)(30) &= 150 \text{ in}^2 &\qquad \text{Right side: } (9.5)(30) &= 285 \text{ in}^2 \\ \text{Back: } (5)(9.5) &=47.5 \text{ in}^2 &\qquad \text{Bottom: } (5)(30) &= 150 \text{ in}^2 &\qquad \text{Left side: } (9.5)(30) &= 285 \text{ in}^2 \end{alignat} $$

Sum of the areas of the 6 faces: $SA = 47.5 + 47.5 + 150+ 150 + 285 + 285 = 965 \text{ in}^2$

Using the surface area formula:

$$ \begin{align} SA &= 2wl + 2lh + 2wh \\ SA &= 2(5)(30) + 2(30)(9.5) + 2(5)(9.5) \\ SA &= 965 \text{ in}^2 \end{align} $$

Students may continue to use the sum of all the areas of the faces to find the surface area which gives the same value as using the surface area formula which is shown below.

Setup of boxDimensions

2 cylinders wide by 1 cylinder high by 3 cylinders long10 in. wide by 9.5 in. high by 15 in. long

$$ \begin{align} SA &= 2wl + 2lh + 2wh \\ SA &= 2(10)(15) + 2(15)(9.5) + 2(10)(9.5) \\ SA &= 775 \text{ in}^2 \end{align} $$

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

coins of silver or gold

step-by-step explanation: