a) constant of variation is 129.6
b) E = 129.6/d^2
c)
Explanation:
Illumination, E varies inversely as distance squared, d²
Writing this mathemeatically:
[tex]\begin{gathered} E\text{ }\alpha\text{ }\frac{1}{d^2} \\ E\text{ = k}\frac{1}{d^2}\text{ (to equate, we will introduce a constant)} \\ E\text{ = }\frac{k}{d^2} \\ \text{where k = cosntant of proportion /variation} \end{gathered}[/tex]
To get k, we will substitute for E and d in the equation
when E = 29.388 lux, d = 2.1 m
[tex]\begin{gathered} E\text{ = }\frac{k}{d^2} \\ 29.388\text{ = }\frac{k}{2.1^2} \\ k\text{ = }29.388(2.1^2) \\ k\text{ = 129.60} \\ \text{Hence, constant of varaition is 129.60} \end{gathered}[/tex]
b) To get the equation that models this situation, we will substitute for k in the equation showing relationship between illumination (E) and distance (d)
[tex]\begin{gathered} E\text{ = }\frac{k}{d^2} \\ E\text{ = }\frac{129.6}{d^2} \end{gathered}[/tex]
c) we need to find the distance when the illumination is 4.614 lux
Using the equation that models the situation:
[tex]\begin{gathered} E\text{ = }\frac{129.6}{d^2} \\ 4.614\text{ = }\frac{129.6}{d^2} \\ 4.614(d^2)\text{ = 129.6} \\ d^2\text{ = }\frac{129.6}{4.614}\text{ = 28.0884} \\ d\text{ = }\sqrt[]{\text{28.0884}} \\ d\text{ = 5.2998} \\ To\text{ the nearest tenth, the distance is 5.3 meters} \end{gathered}[/tex]
