End-of-Unit Assessment

1.

Select all the numbers that are solutions to the equation $x^2=9$ .

A.

81

B.

$\sqrt{81}$

C.

4.5

D.

$\sqrt{9}$

E.

$\sqrt[3]{9}$

F.

3

Answer: D, F

Teaching Notes

Students who select A found the square of 9 instead of the square root. Students who select B squared 9 and then found the square root. Students who select C may be confusing $x^2$ and $2x$ . Students who select E may be confusing a square root with a cube root.

2.

Find the exact length of the segment that joins the points $(\text-7, 2)$ and $(3,5)$ .

A coordinate plane with the origin labeled "O." The x-axis has the numbers negative 7 through 7 indicated. The y-axis has the numbers negative 5 through 5 indicated.

Answer: $\sqrt{109}$

Teaching Notes

Students may graph the points and sketch the segment on the coordinate axes provided, though this is not required. Look out for students making sign errors. Students might use a horizontal length of 4, thinking that

\text-7-3

is 4. Drawing the diagram will make this error much less likely. Students who use signed distances like -10 or -3 and square them incorrectly might wind up with an incorrect answer like

\sqrt{91}

, using

\sqrt{10^2-3^2}

.

3.

Each of the following gives the lengths, in inches, of the sides of a triangle. Which one is a right triangle?

A.

3, 4, 7

B.

$\sqrt{6}$ , 5, 31

C.

$\sqrt5$ , $\sqrt{12}$ , $\sqrt{13}$

D.

$\sqrt{32}$ , $\sqrt{32}$ , 8

Answer:

$\sqrt{32}$ , $\sqrt{32}$ , 8

Teaching Notes

Students who select B or C may have ignored the square root symbols. Students who select A may think that $a^2$ is the same as $2a$ .

4.

What is the decimal expansion of $\frac{16}{9}$ ?

A.

1.7

B.

1.\overline{7}

C.

1.\overline{17}

D.

1.1\overline{7}

Answer: $1.\overline{7}$

Teaching Notes

Students who select A, C, or D may not understand how to represent the repeating part of a decimal or may have made a computation error.

5.

For each number, write the letter of the point that shows its location on the number line.

$\sqrt{3}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt[3]{10}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{12}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{16}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{24}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt[3]{27}$ $\hspace{.1in}\underline{\hspace{.5in}}$

Answer:

$\sqrt{3}$ : J
$\sqrt[3]{10}$ : K
$\sqrt{12}$ : M
$\sqrt{16}$ : N
$\sqrt{24}$ : P
$\sqrt[3]{27}$ : L

Teaching Notes

Sample reasoning students may use to determine the approximate location of each number: $\sqrt{3}$ is between 1 and 2 because 3 is greater than $1^2=1$ but less than $2^2=4$ . $\sqrt[3]{10}$ is a little greater than 2 because 10 is a little greater than $2^3=8$ . $\sqrt[3]{27}$ is equal to 3. $\sqrt{12}$ is between 3 and 4 because 12 is between $3^2=9$ and $4^2=16$ . $\sqrt{16}$ is equal to 4. $\sqrt{24}$ is a little less than 5 since 24 is a little less than $5^2=25$ .

6.

Jada says that $\sqrt{49}$ is irrational because it is a square root. Do you agree with Jada? Explain your reasoning.

Answer:

Sample response:

No, I do not agree with Jada because $\sqrt{49}=7$ since $7^2=49$ , and it can be written as the fraction $\frac71$ . Since it can be written as a fraction, $\sqrt{49}$ is a rational number.

Minimal Tier 1 response:

Work is complete and correct.
Sample: No, because $\sqrt{49}=7$ , which can be written as the fraction $\frac71$ (or equivalent).

Tier 2 response:

Work shows general conceptual understanding and mastery, with some errors.
Sample errors: Response does not explicitly find the value of $\sqrt{49}$ ; response simply states that not all square roots are irrational.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: Explanation shows agreement with Jada; explanation does not include the fact that irrational numbers cannot be written as a fraction or its opposite; no explanation.

Teaching Notes

The explanation that most closely follows the development in this unit is seen in the sample explanation.

7.

Clare has a $\frac12$ -liter bottle full of water. A cone-shaped paper cup has diameter 10 cm and slant height 13 cm as shown. Can she pour all the water into one paper cup, or will it overflow? Explain your reasoning.

(The volume of a cone is $\frac13 \pi r^2h$ and $\frac12$ liter = 500 cubic centimeters.)

A cone, height 13 centimeters, diameter 10 centimeters.

Answer:

The cup will overflow. The radius of the cup is 5 cm since the diameter is 10 cm, and the height of cone is 12 cm because $\sqrt{13^2-5^2}=\sqrt{144}=12$ . The amount of water the cup can hold is the volume of the cone: $V=\frac{\pi}{3}(5)^2(12)=100\pi$ , or approximately 314 cubic centimeters, since $\pi\approx3.14$ . Clare's $\frac12$ -liter bottle has 500 cubic centimeters of water, which is more than the paper cup can hold.

Minimal Tier 1 response:

Work is complete and correct, with complete explanation or justification.
Sample: $\sqrt{13^2-5^2}=12$ . $V=\frac{\pi}{3}(5)^2(12)\approx314$ . The cup can hold about 314 cubic centimeters of water. She cannot pour 500 cubic centimeters into this size cup without it overflowing.

Tier 2 response:

Work shows good conceptual understanding and mastery, with either minor errors or correct work with insufficient explanation or justification.
Sample errors: Omission of units in the final answer; correct mathematical calculations but does not state how much water the cup can hold; use of 10 cm as the radius of the cup.

Tier 3 response:

Work shows a developing but incomplete conceptual understanding, with significant errors.
Sample errors: Incorrect conclusion about water overflowing but volume of cup is correct; correct use of the Pythagorean Theorem to calculate the height of the cup, but no significant further progress; calculation of the surface area of the cup rather than the volume; work involves visual estimation of the radius of the cylinder rather than calculation; calculation for the volume uses the slant height for the height; does not compare volume to the amount of water in a half-liter bottle.

Tier 4 response:

Work includes major errors or omissions that demonstrate a lack of conceptual understanding and mastery.
Sample errors: Work uses neither the Pythagorean Theorem nor the volume formula for a cone; two or more error types under Tier 3 response.

Teaching Notes

In order to get started, students will need to use the slant height and diameter to determine the height of the cone. Also, students will need to use the fact that $\frac12$ a liter is the same as 500 cubic centimeters. Students will also need to recognize that this question is asking for a volume and use the given formula to compute the volume of the cone-shaped cup.