16-20. If a varies directly with the cube of b and inversely with c and b = 2, a = 8 and c = 4. Find a
when b = 6 and c = 12.


Sagot :

Answer:

Step 1: Formula

[tex]a = \frac{kb^{3} }{c}[/tex]

Step 2: Find the constant.

[tex]a = \frac{kb^{3} }{c} \\8 = \frac{k2^{3} }{4} \\4(8 = \frac{k8}{4} )4\\32 = k8\\\frac{32}{8} =\frac{k8}{8} \\4 = k\\k = 4[/tex]

[tex]Constant (k) = 4[/tex]

Step 3: Find a using the value of constant.

[tex]a = \frac{kb^{3} }{c}\\a = \frac{(4)(6)^{3} }{12}\\a = \frac{(4)(216)}{12}\\a = \frac{864}{12} \\a= 72[/tex]

Therefore, a = 72 when b = 6 and c = 12.