Hello everyone! can you me with this problem, and when available, will give out brainliest! thnx!

a shipping company sells two different types of cardboard boxes shaped as regular pyramids. the first box has a square base with side length 30 cm and height 45 cm. the second box has an octagonal base with a perimeter of 120 cm and a distance of 10 cm from the centre of the base to the midpoint of each side. the second box has a height of 35 cm. each box needs to be wrapped in brown paper for shipping. which box requires more paper, and by how much? express your answer in cm2

The first first box needs  600 more brown paper for wrapping

Step-by-step explanation:

Given:

First box:

Square base

Length of the base side =  30 cm

Height of the pyramid = 45 cm

Second Box:

octagonal base

base perimeter =  120 cm

width = 10 cm

Height =  35 cm

To Find:

Which box requires more paper to wrap, and by how much?

Solution:

To Find the amount are brown paper needed to wrap, lets find the surface area of the pyramids

Step: Finding the surface area of the box 1

The First box is square pyramid:

= base square area + 4 x area of side triangle

Base square area:

Area of the square  = side x side

=  30 x 30

=900 centimetre square

Area of triangle

=

=>

=>

=>675 centimetre square

Thus the total brown paper needed for wrapping the first box

= 900 + 675

= 1575  centimetre square--------------------------------(1)

Step 2: Finding the surface area of the box 2

The second box is octagonal pyramid:

= base octagonal area + 8 x area of side triangle

Base octagonal area:

Area of the base of the octagon = 8 x area of one triangle

area of one triangle =  

=

=

=75 centimetre square

Area of the base of the octagon = 8 x 75 = 600 centimetre square

Area of the side triangles  

=  

=

=

= 337.5 centimetre square

Thus the total surface of the octagonal pyramid is  

= 600 +375  

=975 centimetres square -------------------------------(2)

On comparing (1) and(2)

1575 -975 = 600

The first box requires 600 more brown paper for wrapping

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: