Given that y varies jointly with x and with the square root of z, and that y=2 when x=1/8 and z=1/4 , so find i. a formula giving y in terms of x and z, and ii. the value of y when x=3/8 and z=1/9.
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When we are told the y varies jointly with two things then that means y is proportional to the product of these two things, i.e. y is some constant times that product.
So, y = k*x*sqrt(z), or
y = k.x.sqrt(z)
Initial conditions
i.e. y=2 when x=1/8 and z = 1/4, so
2 = k.(1/8).sqrt(1/4)
2 = k.(1/8).(1/2) = k.(1/16)
Therefore, k = 2*16 = 32
Then y = 32.x.sqrt(z)
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The value of y when x=3/8 and z=1/9.
We have y = 32.x.sqrt(z).
Substituting for x = 3/8 and z = 1/9,
y = 32.(3/8).sqrt(1/9)
y = 32.(3/8).(1/3)
y = 32.(1/8)
Answer: y = 4