Solve for x. −23(3x−4)+3x=56

A. x=−196

B. x=−116

C. x=216

D. x=296

Two manned satellites approach one another at a relative speed of 0.400 m/s, intending to dock. The first has a mass of 5.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity in meters per second?

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b