A force F=-2i+j+k acts at point A(1,5,1) on a solid which is free to turn about a fixed point Q(-1,2,1). Find the torque exerted by the force on the body.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

chicken chicken chicken

step-by-step explanation:

there are no multiple choices

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )