can you answer this? i need it as soon as possible.?

Question

if an error of + 0.15, - 0.15 is made in measuring the edge of a cube, what is the approximate error in computed volume?

Answer 1

Error in volume depend on the dimension of edge.
Let say edge of cube = x

for +0.15,
error in volume
= (x+0.15)^3 - x^3
= 0.45x^2 + 0.0675x + 0.003375

for -0.15,
error in volume
= (x-0.15)^3 - x^3
= -0.45x^2 + 0.0675x - 0.003375

if x =1,
error in volume = +0.52, -0.386
if x = 10,
error in volume = +45.678, -44.328

Different edge size determine different error

Answer 2

Multiply the error +-.15 x 3 volume is l x l x l x so the total error could be 0.003375

Answer 3

If you're in a calculus class, use the first derivative of volume.

V=x^3
dV/dx=3*x^2
dV=3*x^2*dx

dx is the error in the linear measurement, and dV is the approximate resulting error in the volume.

Answer 4

+0.15^3, -0.15^3

+0.003375, -0.003375
with 2 significant digits its + or - 0

Answer 5

in the worest case the erorr = +-0.15
ERv = erorr in calculating the volume
ERv= Vexact - Vmeasured
let the length be X ==> Vexact=X^3, V measured in the worest case= (X (+/-) 0.15)^3 = (X^3 (+/-) 0.45X^2 + 0.0675X (+/-) 0.003375
ERv= (+/-)X^2 + 0.0675X (+/-) 0.003375

Answer 6

If you 'know' it is a cube and just measure one edge then it's about +0.52 - 0.39

Answer 7

If y=x^n
and you have an error of +-m in x, then error in y is +-n(m).

So in your case, 3(1.5)=4.5

Answer 8

+/- 0.003375, ie, +/- 0.15 cubed

Answer 9

0.45 +/- i think, since its a cube it will be 0.15 cubed

Answer 10

It would be more or less 0.003375 cubic units