Which is the principal value of \arccos\left(-0.65\right)arccos(−0. 65)

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

table a is wierd

step-by-step explanation:

6*f

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°