When n = 1 first term = -6
n = 2 second term = 0
n = 3 third term = 6
n = 4 4th term = 12
so we have an Arithmetic sequence first term = -6 and common difference = 6
Sum 14 terms = (14/2)[2*-6 + (14-1)*6]
= 462 answer
If the discriminant, b^2 — 4ac is positive (or > 0), then the solution will be two (2) distinct real solutions.
If the discriminant is positive ( > 0), there are 2 real solutions.
If the discriminant is = 0, there is 1 real repeated solution.
If the discriminant is negative (< 0), there are 2 complex solutions (but no real solutions).
Answer:
387 km
Step-by-step explanation:
The length of AC, b can be obtained using the cosine rule ; using the cosine relation:
b² = a² + c² - 2ac CosB
b² = 250² + 185² - 2(250*185) Cos125
b² = 62500 + 34225 - 92500 * −0.573576
b² = 96725 + 53055.820
b² = 149780.82
b = sqrt(149780.82)
b = 387.01527
b = 387 km (nearest whole number)