$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

Step-by-step explanation:

$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

$For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2$

$For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2$

$Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2$

7 0

3 years ago

Describe the different type of polygons

gladu [14]

Convex Polygons

All of its angles are less than 180°.

All of the diagonals are internal.

Concave Polygons

At least one angle measures more than 180°.
At least one of the diagonals is outside the shape of the polygon.

Equilateral Polygons

All sides are equal.

Equiangular Polygons

All angles are equal.

Regular Polygons

They have equal angles and sides

Irregular Polygons

They do not have equal angles and sides.

Types of Polygons based on Number of Sides

Triangle

3 sides.

Quadrilateral

4 sides.

Pentagon

5 sides.

Hexagon

6 sides.

Heptagon

7 sides.

Octagon

8 sides.

Enneagon or Nonagon

9 sides.

Decagon

10 sides.

Hendecagon

11 sides.

Dodecagon

12 sides.

Tridecagon or triskaidecagon

13 sides.

Tetradecagon or tetrakaidecago

14 sides.

Pendedecagon

15 sides.

Hexdecagon

16 sides.

Heptdecagon

17 sides.

Octdecagon

18 sides.

Enneadecagon

19 sides.

Icosagon

20 sides.

3 0

3 years ago

13abc + 2a2d - 14b3c4

Setler [38]

Answer:

13abc + 4ad - 168bc

I don't know if you wrote the problem down right correct me if I'm wrong, please!

3 0

3 years ago

A box of colored crayons contains 16 distinct colors. In how many ways can 14 colors be chosen, assuming that the order of the c

Harrizon [31]

Could you rephrase that? Is it just 14 multiplied by 14?