In triangle abc a=2, c=3, b=95 degrees. find the size of the smallest angle

  A ≈ 32°

Step-by-step explanation:

You are given two sides and the angle between them, so the law of cosines applies. The measure of side b can be found to be ...

  b² = a² + c² -2ac·cos(B)

  b² = 2² +3² -2·2·3·cos(95°) ≈ 14.0459

  b ≈ 3.74778

Then the law of sines can help you find angle A, the angle opposite the shortest side.

  sin(A)/a = sin(B)/b

  A = arcsin(a/b·sin(B)) = arcsin(2/3.74778·sin(95°)) ≈ 32.11°

  A ≈ 32°

The smallest angle is about 32°.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )