find the approximate value of cubic root of 0.026..pls reply fast
Hello Benita, let y = x^1/3
x = 0.027. So y = 0.3
But ∆x = 0.026 - 0.027 = - 0.001
Let us find ∆y
Differentiating ∆ y = 1/3 * x^-2/3 * ∆x
So ∆y = 1/3 * (1/0.09) * -0.001 = -0.1/27 = -1/270
So approximate y = 0.3 - 1/270 = 8/27 = 0.2963
x = 0.027. So y = 0.3
But ∆x = 0.026 - 0.027 = - 0.001
Let us find ∆y
Differentiating ∆ y = 1/3 * x^-2/3 * ∆x
So ∆y = 1/3 * (1/0.09) * -0.001 = -0.1/27 = -1/270
So approximate y = 0.3 - 1/270 = 8/27 = 0.2963