The value of quantity d in which a = 5.7 , b = 4.7 and c is given to be 72 is 0.12.
We are given in the question that the value of a is 5.7m, b is 4.7s, c is 72 m/s and [tex]d=\frac{a^{3}}{cb^{2}}[/tex]
We have to find the value of quantity d.
Putting the values of quantities a, b and c in the given formula of quantity d, we will get,
[tex]d=\frac{(5.7)^{3}}{(72)(4.7)^{2}}[/tex]
First we will find the values of the cube of 5.7 and square of 4.7.
Hence,
[tex](5.7)^{3}=185.193\\\\(4.7)^{2}=22.09[/tex]
Hence, the quantity d will become:-
[tex]d=\frac{185.193}{(72)(22.09)}[/tex]
Simplifying the above equation, we get quantity d as :-
d ≈ 0.116 ≈ 0.12
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