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3 years ago

Pls I need help fast and quick it's urgent!

Mathematics

1 answer:

sammy [17]3 years ago

6 0

Answer:

I might be wrong but I think its the last answer choice.

Step-by-step explanation:

I'm sorry if its wrong :(

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It is currently 0 degrees outside, and the temperature is dropping 4 degrees every hour. The temperature after h hours is -4h&lt

DENIUS [597]

Answer:
A. 3.5
Explanation:
3.5 x 4 = 14

4 0

3 years ago

What is the answer to this definition? To place a dot at the point named by an ordered pair.?

Deffense [45]

The answer is Graph, I hope it helps:)

7 0

3 years ago

#7) Solve 8c - 3c + 3 = 13.

nordsb [41]

Hello!

$\large\boxed{c = 2}$

8c - 3c + 3 = 13

Begin by combining like terms:

5c + 3 = 13

Subtract 3 from both sides:

5c + 3 - 3 = 13 - 3

5c = 10

Divide both sides by 5:

5c/5 = 10/5

c = 2.

6 0

3 years ago

Need help plz help

torisob [31]

Answer:

A. Square for Ape.x

Step-by-step explanation:

8 0

3 years ago

How do I do this?<br>*Look at the directions in the photo*

lora16 [44]

Answer:

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

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