If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

Question

Ethan Adkins · Accepted Answer

#y=21/125# #"the initial statement is "ypropx/z^2# #"to convert to an equation multiply by k the constant"# #"of variation"# #rArry=kxxx/z^2=(kx)/z^2# #"to find k use the given condition"# #y=1/6" when "x=20" and "z=6# #y=(kx)/z^2rArrk=(yz^2)/x=(1/6xx36)/20=3/10# #"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(3x)/(10z^2))color(white)(2/2)|)))# #"when "x=14" and "z=5" then"# #y=(3xx14)/(10xx25)=21/125#

Jace Anderson · Answer

undefined To find y when x = 14 and z = 5, we can use the direct variation and inverse variation formulas.

First, let's find the constant of variation (k) by substituting the given values into the equation y = kx/z^2.

1/6 = k * 20 / 6^2

Simplifying this equation, we get:

1/6 = k * 20 / 36

Now, solve for k:

k = (1/6) * (36/20) = 3/10

Now that we have the value of k, we can find y when x = 14 and z = 5 by substituting these values into the equation:

y = (3/10) * 14 / 5^2

Simplifying this equation, we get:

y = (3/10) * 14 / 25

y = 42/250

Therefore, when x = 14 and z = 5, y is equal to 42/250.