A particle is moving on the x-axis, where x is in centimeters. it has velocity v(x) in cm/s, when it is at the co-ordinate x, given by v(x)=x^3-2x+3 find the acceleration of the particle when it is at the point x=2. express youre answer as a number rounded to the nearest whole number

10cm/s²

Step-by-step explanation:

Acceleration is the change in velocity of an object with respect to time.

Given velocity v(x) to be x³-2x+6

Acceleration = ∆v/∆x

Differentiating the velocity function to get acceleration we have,

Acceleration = dv/dt = 3x²-2

Acceleration of the particle at x = 2 will give;

dv/dt @ x = 2 is 3(2)²-2

= 12-2 = 10cm/s²

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3

h/o/e/s mad

step-by-step explanation:

si si si