Find the measure of AngleJ, the smallest angle in a triangle with sides measuring 11, 13, and 19. Round to the nearest whole degree. 30° 34° 42° 47°

  34°

Step-by-step explanation:

The law of cosines is good for finding angles when only sides are known. We'll use the conventional sides a, b, c, and angles A, B, C. Yes, we know the problem statement calls the smallest angle "J". We trust you can make the translation.

  a² = b² +c² -2bc·cos(A) . . . . . for sides a, b, c and angle A

Solving for the angle, we get ...

  A = arccos((b² +c² -a²)/(2bc))

Filling in the numbers with "a" being the shortest side, we have ...

  A = arccos((13² +19² -11²)/(2·13·19)) = arccos(409/494)

  A ≈ 34.113°

The smallest angle, ∠J, is about 34°.

b. -7,185,024p6q5

step-by-step explanation:

find the 6th term of the expansion of (2p - 3q)11. a. -7,185,024p4q7 c. -7,185p4q7 b. -7,185,024p6q5 d. -7,185p6q5 select the best answer from the choices provided a b c d

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8