f(x,y)= 10x^1/3 y^2/3
Compute (27,64)
so 10(27)^1/3 * (64)^2/3
10 * (27)^1/3 * (64)^2/3
I know that the answer is 480 but I am having a hard time calculating 27^1/3 and 64^2/3.
cube root of 27? 3
cube root of 64, then squared?
64=4*4*4 so the answer is 16
10*3*16=...
Oh, I see. Thank you
To calculate \(27^{1/3}\) and \(64^{2/3}\), you need to understand the concept of exponentiation and fractional exponents. Let's break it down step by step:
1. Start with \(27^{1/3}\):
- The exponent \(1/3\) means you need to find the cube root of \(27\).
- To do that, you can think of a number that, when multiplied by itself three times, equals \(27\). In this case, that number is \(3\).
- So, \(27^{1/3} = 3\).
2. Now, move on to \(64^{2/3}\):
- The exponent \(2/3\) means you need to find the square of the cube root of \(64\).
- First, find the cube root of \(64\). Using the same process as before, you'll see that the cube root of \(64\) is \(4\).
- Then, square that result to get \(4^2\), which equals \(16\).
- So, \(64^{2/3} = 16\).
Now, substitute these values back into the original expression:
\(f(x, y) = 10x^{1/3}y^{2/3} = 10(27)^{1/3} \cdot (64)^{2/3} = 10 \cdot 3 \cdot 16 = 480\).
Therefore, the answer is indeed 480.