If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

Question

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 and solve for k.

Substituting y=20, x=50, and z=5 into the equation y = kx, we get 20 = k * 50. Solving for k, we find k = 20/50 = 2/5.

Now that we have the value of k, we can substitute the given values of x=3 and z=6 into the equation y = k/z^2 and solve for y.

Substituting k=2/5, x=3, and z=6 into the equation y = k/z^2, we get y = (2/5)/(6^2). Simplifying, we find y = 2/5 * 1/36 = 1/90.

Therefore, when x=3 and z=6, y is equal to 1/90.

Eli Assouline · Answer

#y=c/z^2" "->" "y=500/6^2" "->" "y=13 8/9 -> 125/9#

#y=kx " "->" " y=2/5xx3" "->" "y=6/5# As both #x# and #z# are related to #y# we can express them as follows: Given:#" "color(red)( y)" "color(blue)(alpha)" "color(red)( x)" "color(blue)(alpha)" "color(red)( 1/z^2)# Where #alpha# means proportional to. Let #k" and "c# be constants of variation. Then we have: #y=k x=c/z^2# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the values of "k" and "c)# Base condition is that at #y=20 ; x=50 ; z=5# #color(brown)("To determine "k)# #=>y=kx" "->" "20=k(50)# Divide both sides by 50 #20/50=kxx50/50" "# but #50/50=1# #color(brown)(k=20/50=2/5# '........................................................... #color(brown)("To determine "c)# #=>y=c/z^2" "->" "20=c/(5^2)# Multiply both sides by 25 #color(brown)(c=20xx25= 500)# '.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the value of "y" at "x=3)# #y=kx " "->" " y=2/5xx3# #color(blue)( y=6/5)# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the value of "y" at "z=6)# #y=c/z^2" "->" "y=500/6^2# #y=13 8/9 -> 125/9#