Solving For Radius R In The Equation 12 * (3/4) * V = (4/3) * Π * R^3

Hey there, math enthusiasts! Ever stumbled upon an equation that looks like a cryptic puzzle? Well, let's dive into one together and unravel its secrets. Today, we're tackling the equation 12 * (3/4) * V = (4/3) * π * r^3, and our mission, should we choose to accept it, is to isolate and solve for the elusive r – the radius. This equation, at its heart, relates the volume V to the radius r of a sphere, and by manipulating it algebraically, we can express r in terms of V. So, buckle up, grab your algebraic tools, and let's embark on this mathematical adventure!

Decoding the Equation: A Step-by-Step Approach

Our journey begins with the equation 12 * (3/4) * V = (4/3) * π * r^3. The first step in simplifying this equation is to address the numerical coefficients on the left-hand side. We have 12 multiplied by the fraction 3/4. Performing this multiplication, 12 times 3 equals 36, and then dividing by 4 gives us 9. Thus, the left-hand side of the equation simplifies to 9V. This seemingly small simplification is a crucial step, as it reduces the complexity of the equation and makes it easier to manipulate. By combining the numerical terms, we've effectively cleared the path for isolating r on the right-hand side. Remember, in mathematics, simplifying expressions is often the key to unlocking solutions. It's like decluttering a room – once you've cleared the unnecessary items, the important ones become much easier to find. In our case, the 'important item' is the radius, r, and by simplifying the equation, we're getting closer to uncovering its value.

Now, our equation looks cleaner: 9V = (4/3) * π * r^3. To further isolate r we need to tackle the coefficient multiplying _r_³ on the right-hand side, which is (4/3)π. Remember, the golden rule of equation manipulation is to perform the same operation on both sides to maintain balance. Our strategy here is to multiply both sides of the equation by the reciprocal of (4/3), which is (3/4). This will effectively cancel out the (4/3) on the right-hand side. So, we multiply both sides by (3/4). On the left-hand side, we have 9V multiplied by (3/4). This gives us (27/4)V. On the right-hand side, (4/3)π * r^3 multiplied by (3/4) simplifies to π * r^3, as the (4/3) and (3/4) cancel each other out. This step is a perfect example of how using inverse operations can help us isolate the variable we're solving for. It's like using a key to unlock a door – in this case, the 'key' is the reciprocal, and the 'door' is the coefficient that's holding _r_³ captive.

With each step, we're peeling away the layers surrounding r, bringing us closer to our goal. Our equation now stands as (27/4)V = π * r^3. The next hurdle in our quest to isolate r is the presence of π multiplying _r_³. To undo this multiplication, we employ the inverse operation – division. We divide both sides of the equation by π. On the left-hand side, (27/4)V divided by π gives us (27V) / (4π). On the right-hand side, π * r^3 divided by π simplifies to r^3, as the π's cancel each other out. This step highlights the power of inverse operations in algebra. It's like rewinding a tape – we're undoing the multiplication by π to reveal what's underneath. By dividing both sides by π, we've successfully isolated _r_³ on one side of the equation, bringing us one step closer to finding r. It's important to remember that each operation we perform must be applied consistently to both sides of the equation to maintain the equality.

We've arrived at a pivotal point in our journey: (27V) / (4π) = r^3. We've successfully isolated _r_³ but our ultimate goal is to find r, not _r_³. So, how do we undo the cubing operation? The answer lies in the cube root. Just as the square root undoes squaring, the cube root undoes cubing. To isolate r, we need to take the cube root of both sides of the equation. On the left-hand side, we have the cube root of (27V) / (4π). The cube root of 27 is 3, so we can simplify this to 3 times the cube root of V divided by the cube root of 4π. On the right-hand side, the cube root of _r_³ is simply r. This step is like extracting the core essence of a fruit – we're peeling away the outer layers (the exponent of 3) to reveal the inner seed (the value of r). Taking the cube root is a fundamental operation in algebra, allowing us to solve for variables that are raised to the power of 3. By applying this operation to both sides of the equation, we've finally achieved our goal of expressing r in terms of V.

The Grand Finale: Unveiling the Solution

After all the algebraic maneuvering, we've reached the final form of our solution: r = ∛((27V) / (4π)), which can be further simplified to r = (3 * ∛(V)) / ∛(4π). This equation tells us that the radius, r, is equal to the cube root of (27V) divided by (4π). Alternatively, we can express it as 3 times the cube root of V, all divided by the cube root of 4π. This is a significant result because it provides a direct relationship between the volume, V, and the radius, r. If we know the volume of a sphere, we can now use this equation to calculate its radius. This is a powerful tool in various fields, from geometry and physics to engineering and architecture, where understanding the relationship between volume and radius is crucial.

Guys, isn't it amazing how we started with a seemingly complex equation and, step by step, unraveled it to reveal the solution? This journey highlights the beauty and power of algebra – its ability to transform and simplify equations, allowing us to solve for unknown variables. It's like being a detective, piecing together clues to solve a mystery. In this case, the mystery was the value of r, and the clues were the algebraic operations we performed. The final solution, r = (3 * ∛(V)) / ∛(4π), is the culmination of our efforts, the answer we've been seeking. It's a testament to the power of perseverance and the elegance of mathematical reasoning.

Real-World Applications: Where This Equation Shines

Now that we've successfully solved for r, let's take a moment to appreciate the real-world significance of this equation. You might be wondering,