Suppose that y varies directly with x and inversely with z^2, & x=48 when y=8 and z=3. How do you find x when y=12 & z=2?

Question

Eva Adamson · Accepted Answer

undefined To find x when y=12 and z=2, we can use the direct and inverse variation relationship.

First, we need to find the constant of variation.

Given that y varies directly with x, we can write the equation as y = kx, where k is the constant of variation.

Given that y varies inversely with z^2, we can write the equation as y = k/z^2.

To find the constant of variation, we can substitute the given values of x, y, and z into the equation.

When x=48, y=8, and z=3, we have 8 = k(48) and 8 = k/(3^2).

Solving these equations, we find that k = 1/6 and k = 72.

Now, we can use the constant of variation to find x when y=12 and z=2.

Using the equation y = kx, we have 12 = (1/6)x.

Solving for x, we find that x = 72.

Therefore, when y=12 and z=2, x=72.

Hazel Anglin · Answer

#x=32# Equation can be built #y=k*x/z^2# we will find #k# #8=k*48/3^2 => k=(9*8)/48 = 9/6=3/2# now solve for 2nd part #12 = 3/2 * x/2^2 => 12 = (3x)/8# # 4=x/8# #x=32#

Suppose that y varies directly with x and inversely with z^2, &amp; x=48 when y=8 and z=3. How do you find x when y=12 &amp; z=2?

Suppose that y varies directly with x and inversely with z^2, & x=48 when y=8 and z=3. How do you find x when y=12 & z=2?