If y varies inversely as the cube of x and x=2.5 when y=0.015, find y when x=5,?

Question

show work please.

Answer 1

First set up an inverse variation equation using K as your constant of variation:
y = K/x^3

Then substitute in the given values for x and y and solve for K.
0.015 = K/(2.5^3)
0.014 = K/15.625
K = 0.015 * 15.625 = 0. 234275

Now, using the same formula from above, put in your second x value and the K value you just solve for:
y = K/x^3
y = 0.234275/(5^3)
y = 0.234275/125
y = 0.001875

good luck!

Answer 2

The math version of the statement is
y = K/x^3. We can write
0.015 = K/2.5^3
and also y(x=5) = K/5^3
we can form a proportion:
0.015/y = 5^3 K/ 2.5^3 K, which leads to 0.015/y = 8 or y = 0.001875

Answer 3

y = k/(x^3)

0.015 = k/(2.5^3)
k = 0.015 * (2.5^3)
k = 0.015 * 15.625
k = 0.234375
therefore, y = 0.234375/x^3

y = 0.234375/(5^3)
y = 0.001875

Answer 4

y = k/x

y = k/(x^3)

.015 = k/(2.5)^3
k = .015(2.5)^3
k = .234375

y = .234375/(5^3)
y = .001875

Answer 5

y=0.03