Solve the triangle. A = 51°, b = 14, c = 6 A. a ≈ 14.9, C ≈ 28.1, B ≈ 100.9 B. a ≈ 11.2, C ≈ 24.1, B ≈ 104.9 C. a ≈ 14.9, C ≈ 24.1, B ≈ 104.9

(B) has the closest values.

Step-by-step explanation:

Solve the triangle: A = 51°, b = 14, c = 6

A. a ≈ 14.9, C ≈ 28.1, B ≈ 100.9

B. a ≈ 11.2, C ≈ 24.1, B ≈ 104.9

C. a ≈ 14.9, C ≈ 24.1, B ≈ 104.9

Using the cosine rule,

a^2 = b^2+c^2-2bc (cos(A))

= 196+36 - 2(14)(6)cos(51)

=  196+36 - 105.72

= 126.27

a = sqrt(126.27)

= 11.24

using sine rule,

sin(C)/sin(A) = 6/11.24

sin(C) = 6/11.24*sin(51)= 0.41495

C = arcsin(0.41495 = 24.5 degrees, reasonably close to the given value, probably due to the answer used the rounded value of a.

B = 180-51-24.5 =104.5

Out of the given options, only (B) has the correct value of a and C

A. The decrease in Car 1 value is constant at 6000 from year to year. Its value can be described by a linear function.

The decrease in Car 2 value is a constant percentage of its current value from year to year. Its value can be described by an exponential function.

B. The slope of the Car 1 function is -6000, and the value is 32000 in year 1, so the value function for Car 1 can be written as

... f(x) = 32000 -6000(x-1)

... f(x) = 38000 -6000x

The value multiplier for Car 2 from year to year is 27455/32300 = 0.85, so the value function for Car 2 can be written as

... f(x) = 32,300·0.85^(x-1)

... f(x) = 38000·0.85^x

C. There is a significant difference in value after 5 years. Car 2 is worth more than double the value of Car 1 in year 5. That year, their values are ...

... car 1: $8000

... car 2: $16861