In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.00 m/s. it slides a distance of 9.00 m before coming to rest. what is the coefficient of kinetic friction between the puck and the surface? write the answer with 2 decimal places

  about 0.20

Step-by-step explanation:

The speed, acceleration, and distance satisfy the equation ...

  v² = 2ad

So the acceleration is ...

  a = v²/(2d) = (6 m/s)²/(2×9 m) = 2 m/s²

If we take the acceleration due to gravity to be 9.8 m/s², the coefficient of kinetic friction is the ratio of these:

  k = (2 m/s²)/(9.8 m/s²) ≈ 0.20

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!