few_shots: 
Example 1:
Problem/activity:
Task
Materials
One 10-frame per pair of students (see example below)
[IMG]
Counters, 10 per pair of students
Decks of number cards 0-5 (with 4 of each number) or two 0-5 dice
Setup
This activity is done in pairs. The students sit side by side with 10-frame and deck of cards or dice. If they have cards, the cards are shuffled and placed face down in front of them.
Action
Player 1 flips a card or rolls the die. They place that many counters on the 10-frame. Player 2 flips a card or rolls the die. They place that many more counters on the 10-frame next to the counters already on the ten frame. The students count the total together.
A recording sheet can be added when the students have been introduced to equations and number sentences. For those students who can handle higher numbers, a deck of 0-10 cards and two 10-frames can be used.
This game can also be played alone.


Concept/skill: 
Represent addition and subtraction with objects, fingers, mental images, drawings (Drawings need not show details, but should show the mathematics in the problem. (This applies wherever drawings are mentioned in the Standards.)), sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.

Answer: 
Yes. This task involves representing addition with objects such as number cards and dice.

Example 2:
Problem/activity:
For each pair of numbers, determine if the values are equivalent. Explain your reasoning for each one.
a.
$$-0.25$$
and
$$\frac{1}{-4}$$
b.
$$-\left ( \frac{1}{4} \right )$$
and
$$\frac{-1}{-4}$$
c.
$$\frac{-1}{4}$$
and
$$-\frac{1}{4}$$


Concept/skill: 
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.

Answer: 
No. The expressions shown in this problem do not involve multiplying fractions with other fractions, nor whole numbers with fractions.

Example 3:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Concept/skill: 
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

Answer: 
No. This problem does not directly teach how side ratios in right triangles lead to definitions of trigonometric ratios. Instead, it involves solving a trigonometric function.

Now, does the following problem or activity enable students to learn the full intent of the following concept or skill?

shot_info: im_task_001477_pos_0

K

fl_problem_001097_neg_0

7

fl_problem_001894_neg_0

HS_0
standard_description: Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
problem_activity: Activity
In this activity, students find the missing dimensions of cylinders when given the volume and the other dimension. A volume equation representing the cylinder is given for each problem.
Identify students who use these strategies: guess and check, divide each side of the equation by the same value to solve for missing variable, or use the structure of the volume equation to reason about the missing variable
Launch
Arrange students in groups of 2. Give students 2–3 minutes of quiet work time followed by time to share their explanation for the first problem with their partners.
Representation: Internalize Comprehension.
Activate or supply background knowledge of strategies students can use to solve for unknown variables in equations. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Writing, Conversing: MLR1 Stronger and Clearer Each Time.
Use this routine to give students a structured opportunity to revise a response to one or both of the questions. Ask each student to meet with 2–3 other partners in a row for feedback. Provide student with prompts for feedback that will help students strengthen their ideas and clarify their language (e.g., “What did you do first?”, “How do you know
\(h\)
is...?”, and “How did you use the volume formula?”, etc.). Students can borrow ideas and language from each partner to strengthen their final explanation.
Design Principle(s): Optimize output (for explanation)
Student Facing
The volume
\(V\)
of a cylinder with radius
\(r\)
is given by the formula
\(V=\pi r^2h\)
.
The volume of this cylinder with radius 5 units is
\(50\pi\)
cubic units. This statement is true:
\( 50\pi = 5^2 \pi h\)
[IMG]
What does the height of this cylinder have to be? Explain how you know.
The volume of this cylinder with height 4 units is
\(36\pi\)
cubic units. This statement is true:
\(36\pi = r^2 \pi 4\)
[IMG]
What does the radius of this cylinder have to be? Explain how you know.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
Suppose a cylinder has a volume of
\(36\pi\)
cubic inches, but it is not the same cylinder as the one you found earlier in this activity.
What are some possibilities for the dimensions of the cylinder?
How many different cylinders can you find that have a volume of
\(36\pi\)
cubic inches?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Activity Synthesis
Select previously identified students to explain the strategies they used to find the missing dimension in each problem. If not brought up in students’ explanations. Discuss the following strategies and explanations:
Guess and check: plug in numbers for
\(h\)
, a value that make the statements true. Since the solutions for these problems are small whole numbers, this strategy works well. In other situations, this strategy may be less efficient.
Divide each side of the equation by the same value to solve for the missing variable: for example, divide each side of
\(36\pi=r^2 \pi 4\)
by the common factor,
\(4\pi\)
. It’s important to remember
\(\pi\)
is a number that can be multiplied and divided like any other factor.
Use the structure of the equation to reason about the missing variable: for example,
\(50\pi\)
is double
\(25\pi\)
, so the missing value must be 2.


aligns: no