(cube root of 5) * sqrt(5)
--------------------------------- = ?
(cube root of 5^5)
This becomes easier if we switch to fractional exponents:
5^(1/3) * 5^(1/2) 5^(1/3 + 1/2) 5^(5/6)
------------------------ = --------------------- = ------------- = 5^[5/6 - 5/3]
[ 5^5 ]^(1/3) 5^(5/3) 5^(5/3)
Note that 5/6 - 5/3 = 5/6 - 10/6 = -5/6.
1
Thus, 5^[5/6 - 5/3] = 5^(-5/6) = --------------
5^(5/6)
That's the correct answer. But if you want to remove the fractional exponent from the denominator, do this:
1 5^(1/6) 5^(1/6)
---------- * ------------- = -------------- (ANSWER)
5^(5/6) 5^(1/6) 5
100. When rounding to the hundreds place, you must consider the value of the digit in the tens place. If the digit in the tens place is 5 or greater, the digit in the hundreds place increases by 1. If the digit in the tens place is 4 or less the digit in the hundreds place remains the same. In this case, the number in the hundreds place was originally zero.
Im not entirely sure but, if you're on plato answer D is correct
model 1 has a random pattern and is fit for the data
The logarithm of a product is equal to the sum of logarithms:
Drop the exponents:
The logarithm of a number is equal to 1 if the base of the logarithm is the same as the number itself:
