Im stuck on two problems and would really appreciate the help.
1. y varies jointly as a and b, and inversely as the square root of c. y=28 when a=2, b=7, and c=9. Find y when a=4, b=2, and c=4
2. y varies directly as x and inversely as the square of z. y=24 when x=27 and z=3. Find y when x=50 and z=5
y = kab/√c
So, (y√c)/ab = k is constant.
so, you want
(y√4)/(4*2) = (28√9)/(2*7)
Now just find y.
y=kx/z^2
so, you want
(y*25)/50 = (24*9)/27
@Steve Would these be correct?
1. y=64/9
2. y=16
ummm, no.
#1
(y√4)/(4*2) = (28√9)/(2*7)
2y/8 = 84/14
y = 84*4/14 = 24
#2 ok
Sure! I can help you with both of these problems. Let's solve them step by step.
1. For the first problem:
Given that y varies jointly as a and b, and inversely as the square root of c, we can set up the equation as follows:
y = k * (a * b) / √c
To find the value of k, we can substitute the given values of y, a, b, and c:
28 = k * (2 * 7) / √9
28 = k * 14 / 3
k = 28 * 3 / 14
Now that we have the value of k, we can use it to find y when a = 4, b = 2, and c = 4:
y = k * (4 * 2) / √4
Calculate the value of y using the equation above.
2. For the second problem:
Given that y varies directly as x and inversely as the square of z, we can set up the equation as follows:
y = k * (x / z^2)
To find the value of k, we can substitute the given values of y, x, and z:
24 = k * (27 / 3^2)
24 = k * (27 / 9)
k = 24 * 9 / 27
Now that we have the value of k, we can use it to find y when x = 50 and z = 5:
y = k * (50 / 5^2)
Calculate the value of y using the equation above.
By following these steps, you should be able to find the answers to both problems. Let me know if you need further assistance.