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kari74 [83]
3 years ago
14

PLEASE HELP SUPER QUICK WILL GIVE BRAINLIEST AND A TON OF POINTS!!

Mathematics
2 answers:
Scrat [10]3 years ago
5 0

Part a Segment RS Hint: What type of segment is RS? 5 cm     RS is the Radius of circle R the diameter of circle P is 10 cm and circle R and P are congruent and a radius is half of ten so 10 / 2 = 5 cm  

Part c Arc SV Hint: Read the green Arc Measures box on pg. 231. 35 * Angle SRV = 35 * Arc SV = 35 *  Because the degree measure of a minor arc is the measure of it’s central angle   Part d Area of Circle R Leave your answer in terms of π Hint: See the example problem on pg. 192.   25π cm Π r2= π 52 52 = 25 π

25 π cm  


chubhunter [2.5K]3 years ago
4 0
<span> <span><span> <span> Part a </span> <span> Segment RS Hint: What type of segment is RS? </span> <span> 5 cm </span> <span>     RS is the Radius of circle R the diameter of circle P is 10 cm and circle R and P are congruent and a radius is half of ten so 10 / 2 = 5 cm  
<span> <span><span> <span> Part c </span> <span> Arc SV Hint: Read the green Arc Measures box on pg. 231. </span> <span> 35 * </span> <span> Angle SRV = 35 * Arc SV = 35 *  Because the degree measure of a minor arc is the measure of it’s central angle  <span> <span><span> <span> Part d </span> <span> Area of Circle R Leave your answer in terms of π Hint: See the example problem on pg. 192.   </span> <span> 25π cm </span> <span> Π r2= π 52 52 = 25 π
25 π cm  
</span> </span> </span></span> </span> </span> </span></span> </span> </span> </span></span>
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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
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