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Leto [7]
3 years ago
12

Free

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
4 0

Thank you so much!

Have a great day :)

Leona [35]3 years ago
3 0

Answer:

thanks appreciated it

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2b=c-2a
b=(c-2a)/2
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Solve the missing elements for each problem. Use 3.14 for π. Area π r² ; C = π D​
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<h3><u>Answer and Solution</u></h3>

1. Radius = <u>32/2 = 16 inches</u>

Diameter = 32 inches

Circumference = <u>3.14 × 32 = 100.48 inches</u>

Area = <u>3.14 × 256 = 803.84 inches</u>

2. Radius :- <u>40 / 2 = 20 inches</u>

Diameter = 40 inches

Circumference = <u>3.14 × 40 = 125.6 inches</u>

Area = <u>3.14 × 20 × 20 = 1256 inches</u>

3. Radius = <u>28 / 2 = 14 inches</u>

Diameter = 28 inches

Circumference = <u>3.14 × 28 = 87.92 inches</u>

Area = <u>3.14 × 14 × 14 = 615.44 inches</u>

4. Radius = <u>18/2 = 9 inches</u>

Diameter = 18 inches

Circumference = <u>3.14 × 18 = 56.52 inches</u>

Area = <u>3.14 × 9 × 9 = 254.34 inches</u>

5. Radius = 5 inches

Diameter = <u>5 × 2 = 10 inches</u>

Circumference = <u>3.14 × 10 = 31.4 inches</u>

Area = <u>3.14 × 5 × 5 = 78.5 inches</u>

6. Radius = 2 inches

Diameter = <u>2 × 2 = 4 inches</u>

Circumference = <u>3.14 × 4 = 12.56 inches</u>

Area = 3.14 × 2 × 2 = 12.56 inches

<h2>–––––––––☆–––––––––</h2>
8 0
3 years ago
Robin is flying a kite. She ties the 50 foot kite string to the ground. If the kite is
krok68 [10]

\bold{\huge{\orange{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>given </u><u>in </u><u>the </u><u>question </u><u>that</u><u>, </u><u> </u><u>Robin </u><u>is </u><u>flying </u><u>a </u><u>kite</u><u>. </u>
  • <u>She </u><u>ties </u><u>the </u><u>5</u><u>0</u><u> </u><u>foot </u><u>kite </u><u>string </u><u>to </u><u>the </u><u>ground </u><u>.</u>
  • <u>The</u><u> </u><u>kite </u><u>is </u><u>flying </u><u>at </u><u>4</u><u>0</u><u> </u><u>feet </u><u>high </u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>the </u><u>angle </u><u>the </u><u>string </u><u>forms </u><u>with </u><u>the </u><u>group </u><u>that </u><u>is </u><u>angle </u><u>of </u><u>elevation</u><u>. </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>According </u><u>to </u><u>the </u><u>given</u><u> </u><u>question</u><u>, </u>

  • Hypotenuse AC ( Distance of string from the ground) = 50 ft.
  • Perpendicular height AB ( Distance of the kite from the ground) = 40 ft.

<h3><u>Therefore</u><u>, </u></h3>

<u>By </u><u>using </u><u>trigonometric </u><u>ratios</u><u>,</u><u> </u>

{ \bold{\pink{ Sin Φ = }}{\bold{\pink{\dfrac{Perpendicular}{Hypotenuse }}}}

\bold{\red{ Cos Φ = }}{\bold{\red{\dfrac{Base}{Hypotenuse }}}}

\bold{\purple{ Sin Φ = }}{\bold{\purple{\dfrac{Perpendicular}{Base }}}} }

<u>The </u><u>Angle </u><u>of </u><u>elevation </u><u>will </u><u>be </u>

<u>[</u><u> </u><u>The </u><u>angle </u><u>that </u><u>is </u><u>formed </u><u>between </u><u>the </u><u>line </u><u>of </u><u>sight </u><u>and </u><u>base </u><u>of </u><u>the </u><u>triangle </u><u>is </u><u>called </u><u>angle </u><u>of </u><u>elevation </u><u>]</u>

\bold{\pink{Sin Φ = }}{\bold{\pink{\dfrac{AB}{AC }}}}

\sf{ Sin Φ = }{\sf{\dfrac{40}{50 }}}

\sf{ Sin Φ = }{\sf{\dfrac{4}{5}}}

\sf{ Sin Φ = 0.8 }

\bold{\green{  Φ = 45.83° }}

Hence, The measure of the angle the string forms with the ground is 45.83° .

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In short, Your Answer would be Option D) 25

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