If y varies directly as x and inversely as the square of z and if #y=20# when #x=50# and #z=5# how do you find y when #x=3# and #z=6#?

Question

If y varies directly as x and inversely as the square of z and if #y=20#  when #x=50# and #z=5# how do you find y when #x=3# and #z=6#?

Abigail Allen · Accepted Answer

undefined To find y when x=3 and z=6, we can use the direct and inverse variation relationship.

First, we need to determine the constant of variation.

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

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

To find the value of k, we can substitute the given values of y, x, and z into the equation y = kx/z^2.

When y=20, x=50, and z=5, we have 20 = k(50)/(5^2).

Simplifying this equation, we get 20 = 10k.

Solving for k, we find k = 2.

Now, we can use the value of k to find y when x=3 and z=6.

Substituting these values into the equation y = kx/z^2, we have y = 2(3)/(6^2).

Simplifying this equation, we get y = 1/6.

Therefore, when x=3 and z=6, y is equal to 1/6.

Eli Assouline · Answer

#y=5/6# #"the initial statement is "ypropx/z^2# #"to convert to an equation multiply by k the constant"# #"of variation"# #rArry=kxx x/z^2=(kx)/z^2# #"to find k use the given condition"# #y=20" when "x=50" and "z=5# #y=(kx)/z^2rArrk=(yz^2)/x=(20xx25)/50=10# #"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(10x)/z^2)color(white)(2/2)|))# #"when "x=3" and "z=6" then"# #y=(10xx3)/36=30/36=5/6#

If y varies directly as x and inversely as the square of z and if #y=20#&nbsp; when #x=50# and #z=5# how do you find y when #x=3# and #z=6#?&nbsp;&nbsp;

If y varies directly as x and inversely as the square of z and if #y=20# when #x=50# and #z=5# how do you find y when #x=3# and #z=6#?