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1 year ago

After it rained, Albert and his friends saw some worms crawling across the sidewalk. This graph shows the length of the worms.

Mathematics

1 answer:

krek1111 [17]1 year ago

7 0

Answer:

the worm line would be 4 inches long

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Joaquin gets paid every Friday. His paychecks for the past 8 fridays were $324, $245.25, $245.25, $245.25, $324,$324,$245.25 and

zheka24 [161]

Answer:

284.625

Step-by-step explanation:

Arrange the numbers highest to lowest.

Find the middle value in the arrangement.

7 0

2 years ago

PLEASE HELP ME THIS THING IS DUE TOMORROW. 14 feet ≈ ? meters

ddd [48]

14 feet=4.267214 meters

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3 years ago

Hi<br> Someone that is able to do it

VARVARA [1.3K]

I think it’s b

Not for sure but it makes the most sense

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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

2 years ago

HELPPP PLEASE ASAP !!!!!!!!!!

Vikki [24]

Answer:

the triangle flipped over the y axis

6 0

3 years ago

Read 2 more answers