9) If y varies directly as the cube of x and inversely as the square of z, and y = 8 when x = 2 and z = 3. Find y when x = 3 and z = 6.
Answer:
The relationship can be expressed as y = k \cdot \frac{x^3}{z^2}y=k⋅
z
2
x
3
, where kk is the constant of variation. To find kk, plug in the given values: y = 8y=8, x = 2x=2, and z = 3z=3. Then solve for kk:
8 = k \cdot \frac{2^3}{3^2}8=k⋅
3
2
2
3
8 = k \cdot \frac{8}{9}8=k⋅
9
8
k = 8 \times \frac{9}{8} = 9k=8×
8
9
=9
Now that you have kk, use the formula to find yy when x = 3x=3 and z = 6z=6:
y = 9 \cdot \frac{3^3}{6^2}y=9⋅
6
2
3
3
y = 9 \cdot \frac{27}{36}y=9⋅
36
27
y = 9 \cdot \frac{3}{4} = 6.75y=9⋅
4
3
=6.75
Step-by-step explanation:
this problem involves a relationship where yy varies directly with the cube of xx and inversely with the square of zz.
So the equation for this relationship is y = k \cdot \frac{x^3}{z^2}y=k⋅
z
2
x
3
, where kk is a constant that we need to find.
Given that when x = 2x=2 and z = 3z=3, y = 8y=8, we can plug these values into the equation to solve for kk:
8 = k \cdot \frac{2^3}{3^2}8=k⋅
3
2
2
3
Solving this equation gives us k = 9k=9.
Now, to find yy when x = 3x=3 and z = 6z=6, we substitute these values into the equation with the found value of kk:
y = 9 \cdot \frac{3^3}{6^2}y=9⋅
6
2
3
3
Solving this equation gives y = 6.75y=6.75