How do you simplify # 1/4 div 11/12 div 3/4#?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Tony B
Jun 1, 2016

#4/11#

Explanation:

Just to emphasis a point write as:

#(1/4)-:(11/12)-:(3/4)#

The #-:# applies to what is immediately to its right.

The shortcut way to deal with this is to invert (turn upside down) and then multiply.

#=> (1/4)xx(12/11)xx(4/3) = 4/11#

The only purpose of the brackets is to differentiate between the values and the 'operation' applied to them.

I am using the term 'operator' to mean the 'actions' of:
add; subtract; multiply; divide and so on

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