f varies jointly as g2 and h, and f = 50 when g = 4 and h = 2. Find when g = 3 and h = 6
Nice, that's the way I would do it,
To find when g = 3 and h = 6, we need to use the given information.
We are told that f varies jointly as g^2 and h, which means that f = k * g^2 * h, where k is the constant of proportionality.
To find the value of k, we can use the information given: "f = 50 when g = 4 and h = 2."
Substituting these values into the equation, we have:
50 = k * 4^2 * 2
50 = k * 16 * 2
50 = 32k
Now we can solve for k:
k = 50 / 32
k = 1.5625
Now that we have the value of k, we can find f when g = 3 and h = 6:
f = k * g^2 * h
f = 1.5625 * 3^2 * 6
f = 1.5625 * 9 * 6
f = 84.375
Therefore, when g = 3 and h = 6, f is approximately equal to 84.375.
To find when g = 3 and h = 6, we need to use the given information that f varies jointly as g^2 and h.
When we say that f varies jointly as g^2 and h, it means that f is directly proportional to the product of g^2 and h. Mathematically, it can be expressed as:
f = k * g^2 * h
where k is the constant of variation.
To find the value of k, we can use the information given in the problem. It states that f = 50 when g = 4 and h = 2. Substituting these values, we get:
50 = k * 4^2 * 2
50 = k * 16 * 2
50 = k * 32
To solve for k, we can divide both sides of the equation by 32:
50/32 = k
k = 1.5625
Now that we have the value of k, we can substitute it back into the equation to find f when g = 3 and h = 6:
f = 1.5625 * 3^2 * 6
f = 1.5625 * 9 * 6
f = 1.5625 * 54
f ≈ 84.375
Therefore, when g = 3 and h = 6, the value of f is approximately 84.375.
is this correct guys?
F=k•g²•h
50=k(4) ² • 2
50=k•32
K= 50/32 divided by 2
K= 25/16
F= 25/16• (3) ² 6
F= 25/16 • 54/1
F= 1350/16
F= 84.375