Vectors A and B lie in the xy ‑plane. Vector A has a magnitude of 17.1 and is at an angle of 150.5∘ counterclockwise from the +x ‑axis. Vector B has a magnitude of 28.1 and is 205.3∘ from the +x ‑axis. Resolve A and B into components, and express using unit vectors, A =Ax+Ay+Az
B =Bx+By+Bz
where Ax, Ay, Az and Bx, By, and Bz are the calculated values of the x ‑, y ‑, and z ‑components of vectors A and B , respectively.

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

B = -25.41 i ^ -12.0 j ^ + 0 k ^

Explanation:

For this exercise let's use trigonometry by decomposing to vectors

vector A

module 17.1 with an angle of 150.5 counterclockwise.

         Sin 150.5 = / A

         cos 150.5 = Ax / A

         A_{y} = A sin 150.5 = 17.1 sin 150.5

         Aₓ = A cos 1505 = 172 cos 150.5

         A_{y} = 8,420

         Aₓ = -14.870

the vector is

          A = -14.87 i ^ + 8.42 j ^ + 0 k ^

Vector B

          = 28.1 sin 205.3

         Bₓ = 28.1 cos 205.3

         B_{y} = -12.009

          Bₓ = -25.405

the vector is

          B = -25.41 i ^ -12.0 j ^ + 0 k ^

answer: what are you talking about?

explanation:

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2