Answer:
C. c = 64
Step-by-step explanation:
The question is incomplete. Here is the complete question.
What value of c makes the equation true? Assume x greater-than 0 and y greater-than 0
RootIndex 3 StartRoot StartFraction x cubed Over c y Superscript 4 Baseline EndFraction EndRoot = StartFraction x Over 4 y (RootIndex 3 StartRoot y EndRoot) EndFraction
c = 12
c = 16
c = 64
c = 81
Given the function;
[tex]\sqrt[3]{\dfrac{x^3}{cy^4} } = \dfrac{x}{4y\sqrt[3]{y} }[/tex]
We are to find the value of c from the expression.
Step 1: Take the cube of both sides;
[tex](\sqrt[3]{\dfrac{x^3}{cy^4} } )^3= (\dfrac{x}{4y\sqrt[3]{y} })^3\\\dfrac{x^3}{cy^4} = \dfrac{x^3}{(4y)^3(\sqrt[3]{y} )^3}\\\dfrac{x^3}{cy^4} = \dfrac{x^3}{(64y^3)(y)}\\\\\dfrac{x^3}{cy^4} = \dfrac{x^3}{64y^4}\\[/tex]
Step 2: compare the denominator of both sides of the equation;
[tex]\dfrac{x^3}{cy^4} = \dfrac{x^3}{64y^4}\\\\On \ comparing;\\\\cy^4 = 64y^4\\[/tex]
Step 3: Divide both sides by y₄
[tex]\dfrac{cy^4}{y^4} = \dfrac{64y^4}{y^4}\\c = 64\\[/tex]
Hence the value of c is 64. Option C is correct
