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3 years ago

What is the relative size of (how many times as large) a circle's' circumference C to its diameter, d

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

4 0

Answer:

The circumference is pi times larger than the diameter of a circle.

Step-by-step explanation:

The formula of the circumference of a circle is

C = 2(pi)r = (pi)d

C = (pi)d

To find the ratio of the circumference to the diameter, divide both sides of the equation by d.

C/d = pi

The circumference is pi times larger than the diameter of a circle.

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Write the name of the period that has the digits 913

salantis [7]

Nine hundred thirteenths if this is a decimal

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3 years ago

Use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage ofthe total va

Vikentia [17]

Answer:

Step-by-step explanation:

The coefficient of determination = $r^{2} = 0.885^{2}$ = 0.7832

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3 years ago

Find VALUE X<br> PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

amm1812

Answer:

Step-by-step explanation:

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3 years ago

I dont understand these problems |×| - 1 < 5 , 3x + 4 <-2, 5x/7 -2=1, 9(2x+3)=10

Marta_Voda [28]

$|x|-1$

3 0

4 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