Question

Simplify the square root of 5 times the cube root of 5

Answer 1

Simplify the square root of 5 times the cube root of 5.

five to the five sixths power
five to the one sixth power
five to the two thirds power
five to the seven sixths power

Answer 2

The number of users of the internet in a town increased by a factor of 1.01 every year from 2000 to 2010. The function below shows the number of internet users f(x) after x years from the year 2000:

f(x) = 3000(1.01)x

Which of the following is a reasonable domain for the function?

0 ≤ x ≤ 10
2000 ≤ x ≤ 2010
0 ≤ x ≤ 3000
All positive integers

Answer 3

To simplify the expression, we can rewrite it as a single radical:

√5 * 5^(1/3)

Since both terms have 5 as a common factor, we can factor it out:

5 * (√5 * (5^(1/3) / 5))

Now, let's simplify the cube root of 5:

5 * (√5 * 5^(1/3) / 5)
= 5 * (√5 * (∛5^3 / 5))
= 5 * (√5 * (∛125 / 5))

The cube root of 125 is 5 because 5 * 5 * 5 = 125.

Now we have:

5 * (√5 * (5 / 5))
= 5 * (√5 * 1)
= 5 * √5

Therefore, the simplified form of the expression is 5√5.

Answer 4

To simplify the square root of 5 times the cube root of 5, we can express both terms with the same index.

The square root of 5 can be written as 5^(1/2).
The cube root of 5 can be written as 5^(1/3).

Combining these expressions, we have:

5^(1/2) * 5^(1/3)

To simplify this, we can combine the exponents:

5^(1/2 + 1/3) = 5^(3/6 + 2/6) = 5^(5/6)

Therefore, the simplified expression is "five to the five-sixths power."

Answer 5

The reasonable domain for the function f(x) = 3000(1.01)^x is determined by the context of the problem.

Since the function represents the number of internet users after x years from the year 2000, a reasonable domain for x would be the range of years mentioned in the problem statement, which is from 2000 to 2010.

Therefore, the reasonable domain for the function is: 2000 ≤ x ≤ 2010.