Radicals and Exponents: Mastery Test 9th grade Algebra

Answer 1

1. Simplify the following expression:

$\sqrt{25} \cdot \sqrt{9}$

Solution:

$\sqrt{25} = 5$

$\sqrt{9} = 3$

$5 \cdot 3 = 15$

So, $\sqrt{25} \cdot \sqrt{9} = 15$

2. Evaluate the following expression:

$2^3 \cdot 4^2$

Solution:

$2^3 = 2 \cdot 2 \cdot 2 = 8$

$4^2 = 4 \cdot 4 = 16$

$8 \cdot 16 = 128$

So, $2^3 \cdot 4^2 = 128

3. Simplify the following expression:

$\sqrt{64} - \sqrt{16}$

Solution:

$\sqrt{64} = 8$

$\sqrt{16} = 4$

$8 - 4 = 4$

So, $\sqrt{64} - \sqrt{16} = 4$

4. Evaluate the following expression:

$5^2 \div 25$

Solution:

$5^2 = 5 \cdot 5 = 25$

$25 \div 25 = 1$

So, $5^2 \div 25 = 1$

5. Simplify the following expression:

$\sqrt{49} + \sqrt{81}$

Solution:

$\sqrt{49} = 7$

$\sqrt{81} = 9$

$7 + 9 = 16$

So, $\sqrt{49} + \sqrt{81} = 16

6. Evaluate the following expression:

$3^4 \div 3^2$

Solution:

$3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81$

$3^2 = 3 \cdot 3 = 9$

$81 \div 9 = 9$

So, $3^4 \div 3^2 = 9

7. Simplify the following expression:

$4\sqrt{125} - 2\sqrt{27}$

Solution:

$4\sqrt{125} = 4 \cdot 5 = 20$

$2\sqrt{27} = 2 \cdot 3 = 6$

$20 - 6 = 14$

So, $4\sqrt{125} - 2\sqrt{27} = 14$

8. Evaluate the following expression:

$2^5 \cdot 2^{-3}$

Solution:

$2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$

$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$

$32 \cdot \frac{1}{8} = 4$

So, $2^5 \cdot 2^{-3} = 4

9. Simplify the following expression:

$\sqrt{16 \cdot 9}$

Solution:

$\sqrt{16 \cdot 9} = \sqrt{144} = 12$

So, $\sqrt{16 \cdot 9} = 12$

10. Evaluate the following expression:

$3^3 \div 3^4$

Solution:

$3^3 = 3 \cdot 3 \cdot 3 = 27$

$3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81$

$27 \div 81 = \frac{1}{3}$

So, $3^3 \div 3^4 = \frac{1}{3}$