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: 
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.

Answer: 
No. Though this task involves addition, it does not involve any word problems.

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 problem or activity shown below enable students to learn the full intent of the following concept or skill?

shot_info: im_task_001477_neg_0

K

fl_problem_001097_neg_0

7

fl_problem_001894_neg_0

HS_0
standard_description: Interpret multiplication as scaling (resizing), by:
- Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n×a)/(n×b) to the effect of multiplying a/b by 1.
- Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
problem_activity: Narrative
The purpose of this activity is for students to find the missing value in equations involving a whole number and a fraction. They find missing factors or products to make equations true. Many of the problems encourage students to think in steps, interpreting a fraction
\(\frac{a}{b}\)
in terms of division by the denominator and multiplication by the numerator.
Representation: Internalize Comprehension
. Synthesis: Invite students to identify which details were most important to solve the problems. Display the sentence frame: “The next time I solve multiplication problems of a whole number by a fraction, I will pay attention to . . . .“
Supports accessibility for: Conceptual Processing, Memory
Launch
Groups of 2
Activity
5–8 minutes: independent work time
1–2 minutes: partner discussion
Monitor for:
partners who use different strategies.
students who are challenged but persevere and arrive at a viable solution.
Student Facing
Fill in the blanks to make each equation true. Be prepared to explain your reasoning.
\(\frac{1}{3} \times 18 =\underline{\hspace{0.7cm}}\)
\(\frac{7}{9} \times \underline{\hspace{0.7cm}} = \frac{21}{9}\)
\(\frac{1}{15} \times \underline{\hspace{0.7cm}} = 2\)
\(9 \times  6 \frac{2}{3}  = \underline{\hspace{0.7cm}} \)
\(14 \frac {99}{100} \times 10 = \underline{\hspace{0.7cm}} \)
\(7 \frac {3}{5} \times 6 = \underline{\hspace{0.7cm}}\)
\(4 \times 6 \frac {9}{10} = \underline{\hspace{0.7cm}}\)
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Advancing Student Thinking
If students do not get the correct solution for the missing factor problems, ask “How would you represent this expression with a diagram?”
Activity Synthesis
Ask previously selected students to share their responses.
“What was different about you and your partner's strategies?”
“Which problem did you find most challenging and why?”
“Which problem made the most sense to you and why?”


aligns: yes