Unit 1 Pretest

Both numbers are rational because 16 is a perfect square and a perfect cube.

Both numbers irrational because the decimal form of each number does not repeat or terminate.

The decimal form of $$\sqrt[]{16}$$ is irrational because the number does not repeat or terminate. The decimal form of $$\sqrt[3]{16}$$ is rational because the number terminates.

The decimal form of $$\sqrt[3]{16}$$ is irrational because the number does not repeat or terminate. The decimal form of $$\sqrt[]{16}$$ is rational because the number terminates.

Frankie is correct because the $$\sqrt[]{20}$$ is 10 which is a rational number.

Alejandro is correct because the $$\sqrt[]{20}$$ falls between 4 and 5 and the decimal terminates.

Both students are correct because $$\sqrt[]{20}$$ is a rational number.

Neither student is correct because $$\sqrt[]{20}$$ is an irrational.

The decimal expansion $$37.\overline{2}$$ represents $$37\frac{2}{10}$$

The decimal expansion 37. $$37.\overline{20}$$ represents $$37\frac{2}{9}$$

The decimal expansion $$37.0\overline{2}$$ represents $$37\frac{2}{99}$$

The decimal expansion $$37.00\overline{2}$$ represents $$37\frac{2}{900}$$

$$20\div4=5$$

$$20\div5=4$$

$$20-16=4$$

$$25-20=5$$

$$4\times4=16$$

$$4\times5=20$$

$$20\div4=5$$

$$36\div9=4$$

$$\frac{3}{2}$$

$$\frac{1}{2\left(3\right)}$$

$$\frac{1}{2\left(3^5\right)}$$

$$\frac{5^4}{2\left(3^5\right)}$$

Both students incorrectly applied the power of a

quotient for the base of 3.

Chandler applied the power of a quotient for the base of 3 first, while Grace applied it last.

Grace divided the exponents of the base 3 when the exponents should have been subtracted.

Chandler divided the exponents of the base 3 when the exponents should have been subtracted.

$$x^2=9$$

$$x^2=6$$

$$x^3=9$$

$$x^3=27$$

$$k=\sqrt[]{361}=19$$

$$k=\sqrt[]{361}=\pm19$$

$$k=\pm\sqrt[]{361}=19$$

$$k=\pm\sqrt[]{361}=\pm19$$

4

$$\frac{1}{4}$$

-2

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