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thnx for the help

## Homework Statement

The length of a side of the square base of a pyramid is

*x*metres.

The vertical height is

*y*metres.

Both measurements are correct to the nearest metre.

question) Find an expression for the difference between the upper bound and the lower bound of the volume of the pyramid.

Give your answer in its simplest form.

## Homework Equations

volume of pyramid = (1/3)base x height

## The Attempt at a Solution

**my notation: (lb) means lower bound of, so (lb)y would mean the lower bound of y, and so on.**

**p.s. I know some (well alot) are redundant but I stuck them in to make the workings easier to read. Also, i should really learn Latex but I havnt got round to yet so please try to read my non-latex stuck lol.**

(lb)x = x - (1/2) x (1/10)x

(lb)x = x - (1/20)x

(lb)x = (19/20)x

therefore: (ub)x = (21/20)x

(lb)y = (19/20)y

(ub)y = (21/20)y

(lb)base = (lb)x^2

(lb)base = (19x/20)^2

therefore: (ub)base = (21x/20)^2

(lb)volume = (1/3) x (lb)b x (lb)h

v = (1/3) x (19x/20)^2 x (19y/20)

(ub)volume = (1/3) x (21x/20)^2 x (21y/20)

difference = (ub)vol - (lb)vol

d = (1/3) x (21x/20)^2 x (21y/20) - (1/3) x (19x/20)^2 x (19y/20)

d = (21y x (21x/20)^2)/60 - (19y x (19x/20)^2) /60

d = (21y x (21x/20)^2 - 19y x (19x/20)^2) /60

is this the simplest form? seem rather complex to me.

I tried factorising the brackets and doing all sorts but I couldn't get it any cleaner than this.

thnx for your help