Question 692536
y varies jointly as m and the square root of n and inversely as p.
y = {{{(m*sqrt(n))/p}}}
:
y=15 when m=2,n=1, and p=6.
let k = the constant of variation
{{{(k(m*sqrt(n)))/p}}} = y
then replace y, m, n, and p; find k
{{{(k(2*sqrt(1)))/6}}} = 15
{{{(2k)/6}}} = 15
2k = 15 * 6
2k = 90
k = 90/2
k = 45
then
y = {{{(45(m*sqrt(n)))/p}}} is our equation
:
find y when m=3, n=4, and p=10
y = {{{(45(3*sqrt(4)))/10}}}
y = {{{(45(6))/10}}}
y = {{{270/10}}}
y = 27