Find the angle of the sector given the radius and the area of the sector.

Mathematics

1 answer:

gregori [183]3 years ago

5 0

Answer:

270 degrees

Step-by-step explanation:

$Area \: of \: the \: sector \\ \\ = \frac{ \theta}{360 \degree} \times \pi {r}^{2} \\ \\ 339.3 = \frac{ \theta}{360 \degree} \times 3.14 {(12)}^{2} \\ \\ 339.3 = \frac{ \theta}{360 \degree} \times 3.14 \times 144 \\ \\ \theta = \frac{339.3 \times 360 \degree}{3.14 \times 144} \\ \\ \theta = \frac{122,148\degree}{452.16} \\ \\ \theta= 270.143312 \degree \\ \\ \theta \approx \: 270\degree$

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David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o

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Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :( $b^4,b^2,b^4$ )

$b^4$ = b × b × b × b

$b^2$ = b × b

$b^4$ =b × b × b × b

GCF of variables :( $b^4,b^2,b^4$ ) is $b^2$

3) GCF of $c^3$ and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.