what a strange notation. I'd have said
cuberoot(3r-6) = 3
so,
3r-6 = 3^3 = 27
3r = 33
r = 11
cuberoot(3r-6) = 3
so,
3r-6 = 3^3 = 27
3r = 33
r = 11
Step 1: Start with the equation: (index of 3)square root of 3r-6 = 3.
Step 2: Raise both sides of the equation to the power of 3 to remove the index 3 on the left side: [(index of 3)square root of 3r-6]^3 = 3^3.
Step 3: Simplify the left side of the equation: 3r - 6 = 27.
Step 4: Add 6 to both sides of the equation: 3r = 33.
Step 5: Divide both sides of the equation by 3 to solve for "r": r = 11.
Therefore, the value of "r" that satisfies the equation is 11.
Step 1: Square both sides of the equation to eliminate the square root:
(√(3r-6))^2 = 3^2
This simplifies to:
3r-6 = 9
Step 2: Add 6 to both sides of the equation to isolate the term with 'r':
3r - 6 + 6 = 9 + 6
This simplifies to:
3r = 15
Step 3: Divide both sides of the equation by 3 to solve for 'r':
3r / 3 = 15 / 3
This simplifies to:
r = 5
So, the value of r that satisfies the equation √(3r-6) = 3 is r = 5.