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3 years ago

What is the measure of the minor arc GH?

Mathematics

1 answer:

Ivahew [28]3 years ago

4 0

Answer:

a minor is a arc smaller than a semi circle if that helps

Step-by-step explanation:

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VashaNatasha [74]

Answer:

Step-by-step explanation:

As it is a list of prime number, the sequence continues with 29 being the next prime number after 23.

6 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

1,2,5,6,7,11 slope and y-intercept

enot [183]

I’m sorry I can’t help ypu

7 0

3 years ago

Pls help with number 4b

Leya [2.2K]

Answer:

where is the hint

Step-by-step explanation:

you might want to try reading the hint

7 0

3 years ago

A highway map of Ohio has a coordinate grid superimposed on top of the state. Springfield is at point (1, –4) and Columbus is at

Vedmedyk [2.9K]

The midpoint, where the rest area is, will just be the average of the coordinates of Springfield and Columbus.

Rest Area=((1+7)/2, (-4+1)/2)

Rest Area=(4, -1.5)

The distance between the two cities can be found using the Pythagorean Theorem, which used in the manner is often referred to as the "distance" formula between two points.

d^2=(x2-x1)^2+(y2-y1)

d^2=(7-1)^2+(1--4)^2

d^2=36+25

d^2=61

d=√61 and we are told that each unit is equal to 5.38mi so the distance from Springfield to Columbus is:

D=5.38√61 mi

D≈42.02mi (to nearest hundredth of a mile)

3 0

4 years ago