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3 years ago

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

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

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7 0

2 years ago

Read 2 more answers

Find the number to add to x^2-18x

pentagon [3]

Answer:

A. add 81; $(x-9)^2$

Step-by-step explanation:

The perfect square formula

$(a-b)^2=a^2-2ab+b^2$

Rewrite expression $x^2-18x$ as follows:

$x^2-2\cdot 9\cdot x$

So, to form a perfect square, $+9^2$ is missed.

When we add $9^2=81,$ we will get

$x^2-2\cdot 9\cdot x+9^2=(x-9)^2$

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3 years ago

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andrew11 [14]

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Step-by-step explanation:

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3 years ago

Find the midpoint between (4,-1) and (3,2)

zepelin [54]

Midpoint has form of,

$M(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

We can insert the data,

$M(\dfrac{4+3}{2},\dfrac{-1+2}{2})$

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r3t40

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3 years ago

How do I do this? *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