Suppose that x varies directly as y ^2 and inversely as z, and x is 5 when y is 3 and z is 3. What is z when x is 10 and y is 3?
You can do it without finding k explicitly. You do know that
x = ky^2z, so x/(y^2z) = k, a constant. Thus, you just need to find z such that
10/(9z) = 5/(9*3)
or, even more simply in this case, since y has not changed, if x is twice as big, so is z.
To solve this problem, we need to find the equation that describes the relationship between x, y, and z.
The problem states that x varies directly as y^2, which means that x is directly proportional to the square of y. Mathematically, we can write this as:
x = k * y^2
where k is a constant of proportionality.
The problem also states that x varies inversely as z, which means that x is inversely proportional to z. This can be represented as:
x = k * (1/z)
Using both of these statements together, we can write the equation for x as:
x = k * y^2 * (1/z)
Now, we can plug in the given values to find the value of k.
When x is 5, y is 3, and z is 3:
5 = k * (3^2) * (1/3)
5 = 9k/3
15 = 9k
k = 15/9
k = 5/3
Now that we have the value of k, we can use it to find the value of z when x is 10 and y is 3.
Plugging in the new values:
10 = (5/3) * (3^2) * (1/z)
10 = (5/3) * 9 * (1/z)
10 = 15/z
To solve for z, we can cross-multiply:
10z = 15
Dividing both sides by 10:
z = 15/10
z = 3/2
Therefore, when x is 10 and y is 3, z is equal to 3/2.
"x varies directly as y ^2 and inversely as z"
----> x = k(y^2 * 1/z)
repeat the method I used for your other two problems.