The length of a rectangle is increasing at a rate of 9 cm/s and its width is increasing at a rate of 7 cm/s. When the length is 12 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

129 m/s^2

Step-by-step explanation:

The length of a rectangle is increasing at a rate of 9m/s

dL/dt = 9m/s

The width is increasing at a rate of 7m/s

dw/dt= 7m/s

The formular for solving the area of a rectangle is length × width

Therefore, to calculate how fast the rectangle is increasing we will apply the product rule

dA/dt= L × dw/dt + W × dl/dt

= 12×7 + 5×9

= 84+45

= 129m/s^2

Hence the area of the rectangle is increasing at 129m/s^2

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

explanation:

step-by-step explanation: