Let's go through each of the problems one by one, providing detailed solutions for all the questions in the image.

Nos. 31-35: Evaluate each expression or state that it is not a real number.

31) $\sqrt{-16}$

The square root of a negative number is not a real number, as square roots of negative numbers are complex.

$\sqrt{-16} = 4i$ where $i$ is the imaginary unit, $i = \sqrt{-1}$.

Answer: $4i$ (Not a real number).

32) $(-32)^{\frac{1}{5}}$

This is the fifth root of -32. Since we are dealing with an odd root (5th root), the result is real.

$(-32)^{\frac{1}{5}} = -2$

Answer: $-2$ (Real number).

33) $\sqrt[3]{0}$

The cube root of 0 is always 0, as any number raised to any power that gives 0 must be 0 itself.

$\sqrt[3]{0} = 0$

Answer: $0$ (Real number).

34) $\sqrt{-1}$

This is the square root of -1, which is not a real number. By definition:

$\sqrt{-1} = i$

Answer: $i$ (Not a real number).

35) $(27)^{\frac{2}{3}}$

This can be broken down as follows: $(27)^{\frac{2}{3}} = \left( \sqrt[3]{27} \right)^2$ First, find the cube root of 27: $\sqrt[3]{27} = 3$ Now square the result: $3^2 = 9$

Answer: $9$ (Real number).

Nos. 36-40: Simplify (if possible) the radical expressions.

36) $\sqrt{2} - 5\sqrt{2}$

Both terms involve $\sqrt{2}$, so we can treat them as like terms:

$\sqrt{2} - 5\sqrt{2} = (1 - 5)\sqrt{2} = -4\sqrt{2}$

Answer: $-4\sqrt{2}$.

37) $\sqrt{16x^3y}$

We can simplify the square root by breaking it into parts:

$\sqrt{16x^3y} = \sqrt{16} \cdot \sqrt{x^3} \cdot \sqrt{y}$

Now simplify each part: $\sqrt{16} = 4, \quad \sqrt{x^3} = x^{\frac{3}{2}} = x \sqrt{x}, \quad \sqrt{y} = \sqrt{y}$

Putting everything together: $4x\sqrt{xy}$

Answer: $4x\sqrt{xy}$.

38) $3\sqrt{5} - 2\sqrt{5} + 7\sqrt{5}$

All terms involve $\sqrt{5}$, so we can combine them like like terms:

$3\sqrt{5} - 2\sqrt{5} + 7\sqrt{5} = (3 - 2 + 7)\sqrt{5} = 8\sqrt{5}$

Answer: $8\sqrt{5}$.

39) $\sqrt[3]{-81x^6y^8}$

Break this down by taking the cube root of each part separately: [ \sqrt[3]{-81x^6y^8} = \sqrt[3]{-81} \cdot \sqrt[3]{x^6} \cdot \sqrt[3]{y^8} ]

Simplify each part: [ \sqrt[3]{-81} = -\sqrt[3]{81} = -3, \quad \sqrt[3]{x^6} = x^2, \quad \sqrt[3]{y^8} = y^{\frac{8}{3}} ]

Putting everything together: $-3x^2y^{\frac{8}{3}}$

Answer: $-3x^2y^{\frac{8}{3}}$.

40) $\sqrt[5]{-32x^{10}y^8}$

This can be simplified by taking the fifth root of each part separately:

[ \sqrt[5]{-32x^{10}y^8} = \sqrt[5]{-32} \cdot \sqrt[5]{x^{10}} \cdot \sqrt[5]{y^8} ]

Simplify each part: [ \sqrt[5]{-32} = -2, \quad \sqrt[5]{x^{10}} = x^2, \quad \sqrt[5]{y^8} = y^{\frac{8}{5}} ]

Putting everything together: $-2x^2y^{\frac{8}{5}}$

Answer: $-2x^2y^{\frac{8}{5}}$.

Do you have any questions or need further clarification on any of the steps?

Here are some related questions to explore:

1. How do you simplify higher-order roots of negative numbers?
2. Why is the square root of a negative number not a real number?
3. What are the rules for combining like radicals?
4. How do cube roots behave differently from square roots?
5. What is the significance of rational exponents in radical expressions?

Tip: When simplifying radicals, always break the expression down into parts and simplify each part individually before combining them!