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Harman [31]
3 years ago
15

Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

AWARDING BRAINLIEST!!!

Mathematics
2 answers:
liberstina [14]3 years ago
6 0

Answer:

40π in²

Step-by-step explanation:

The area (A) of sector ODC is calculated as

A = area of circle × fraction of circle

   = πr² × \frac{72}{360}

   = π × 10² × \frac{1}{5} = \frac{100\pi }{5} = 20π cm²

Sector OAB is congruent to sector ODC, thus

area of shaded regions = 2 × 20π = 40π in²

EastWind [94]3 years ago
5 0

Answer:

126 in²

Step-by-step explanation:

10.10.0,628.2=126

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1 1/+6

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9/16 + 8/16=17/16=1 1/16

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In JKL, k=4.1 cm,j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .
spayn [35]

Answer:

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the  nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

\frac{k}{sin(K)}=\frac{j}{sin(J)}   \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)}  \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o  \\

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sveticcg [70]

Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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  • Left to Right

Equality Properties

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  • Addition Property of Equality
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Step-by-step explanation:

<u>Step 1: Define</u>

5x = 2

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Divide 5 on both sides:                    x = 2/5

<u>Step 3: Check</u>

<em>Plug in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                     5(2/5) = 2
  2. Multiply:                                2 = 2

Here we see that 2 does indeed equal 2.

∴ x = 2/5 is the solution to the equation.

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