Sorry - me again. Can you please show work for the following problems on page 102:

#2, #6, #7

Thanks

#2

(z^4)/z = z^3

(y^3)/(y^2) = y

48/12 = 4

so we have 4zy

#6

2^3 = 8
6^2 = 36
18 = 18
16^(1/2) = 4

so we have 8/4 * 36/18 = 2*2 = 4


#7

15^3 = 5^3 * 3^3
8 = 2^3
2^2 = 2^2
6^3 = 2^3 * 3^3

so we have (5^3)(3^3)(2^3) / ((2^2)(2^3)(3^3) )

= (5^3) / ((2^2) )

= 125/4


- Veritas Help

In problem number 6 you stated that 16^1/2= 4. Can you explain this? More specifically how do we solve when a whole number is raised to a fraction? In addition, how do you arrive at 8/4 and 36/18? Which ultimately leads to the final answer of 4.

If 16 is raised to the 1/2, it's the same as taking the square root.
If it's raised to the 1/3, it's the same as taking the cube root.
1/4 means we take the fourth root, etc.

See page 104 in Math Essentials for more on this.

As for the way the steps work together:

We start with numerator = 2^3 * 6*2 and denominator = 18 * 16^(1/2)

2^3 becomes 8
6^2 becomes 36
18 stays 18
16^(1/2) becomes 4

Now, our numerator is 8 * 36 and our denominator is 18 * 4
The 8 and 4 cancel, leaving a 2 on top, and the 36 and 18 cancel, leaving another 2 on top.
This gives us 4 as the final answer.

Veritas Prep