Question

1. z varies directly with ( x^2 ) and inversely with ( y ). when ( x = 2 ) and ( y = 4 ), ( z = 3 ). what is the value of ( z ) when ( x = 4 ) and ( y = 9 )?

options: ( \frac{16}{3} ), ( \frac{2}{3} ), 24, ( \frac{8}{3} )

Explanation:

Step1: Define variation formula

$z = k\frac{x^2}{y}$

Step2: Calculate constant k

$3 = k\frac{2^2}{4} \Rightarrow 3 = k\frac{4}{4} \Rightarrow k=3$

Step3: Compute z at x=4,y=9

$z = 3\times\frac{4^2}{9} = 3\times\frac{16}{9} = \frac{48}{9} = \frac{16}{3}$? Wait, no—wait, original problem: when x=2,y=4,z=3? Wait, no, wait the problem says z varies directly with x² and inversely with y. Let's recheck: if x=2,y=4,z=3: $3 = k*(4)/4 \Rightarrow k=3$. Then x=4,y=9: $z=3*(16)/9 = 16/3 ≈5.333$, but the options include 24, 8/3, etc. Wait maybe I misread the initial values? Oh wait, maybe the initial z is different? Wait no, the problem says: "z varies directly with x² and inversely with y. When x=2 and y=4, z=3. What is the value of z when x=4 and y=9?" Wait 3*(16)/9=16/3, which is an option (16/3). Oh right! So the answer is 16/3.

Wait correction:

Step1: Define variation

$z = k\frac{x^2}{y}$

Step2: Find k

$3 = k\frac{2^2}{4} \Rightarrow k=3$

Step3: Calculate z

$z=3\times\frac{4^2}{9}=3\times\frac{16}{9}=\frac{16}{3}$

Step1: Set up variation formula

$z = k\frac{x^2}{y}$

Step2: Solve for constant k

$3 = k\frac{2^2}{4} \Rightarrow k=3$

Step3: Compute z with x=4,y=9

$z = 3\times\frac{4^2}{9} = \frac{16}{3}$

Answer:

24