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 and division and of fractions to multiply and divide rational numbers.

Answer: 
Yes. This problem involves understanding fractions and division to divide rational numbers, such as 1, -4, 4, and -1. This problem also involves an understanding of multiplication: if p and q are integers, then -(p/q) = (-p)/q = p/(-q).

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


Concept/skill: 
Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.

Answer: 
Yes. An initial step when solving for x in this problem requires the use of the addition formula for cosine on the left side of the equation.

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_pos_0

7

fl_problem_001894_pos_0

HS_0
standard_description: Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.
problem_activity: Activity
The purpose of this activity is to extend the definition of exponents to include negative exponents. The meaning of zero exponents is also revisited. When students use tables to explore expressions with negative exponents, they are noticing and making use of structure (MP7).
Launch
Arrange students in groups of 2. Tell students to work on their own and then share their reasoning with their partner after each has had a chance to complete the problem. Follow with a whole-class discussion.
Student Facing
Complete the table.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "col"}}, {"_attributes": {"scope": "col"}, "_value": "exponential form"}, {"_attributes": {"scope": "col"}, "_value": "number form"}, {"_attributes": {"scope": "col"}, "_value": "calculations"}]}]}], "tbody": [{"tr": [{"td": [{}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "1145"}, "_value": "\\(2^5\\)"}]}, {}, {}]}, {"td": [{}, {}, {"_value": "16"}, {}]}, {"td": [{"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "25037"}, "_value": "\\(\\frac{2^4}{2}=2^{4-1}=2^3\\)"}]}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "1932"}, "_value": "\\(2^3\\)"}]}, {}, {}]}, {"td": [{"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "25045"}, "_value": "\\(\\frac{2^3}{2}=2^{3-1}=2^2\\)"}]}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "1131"}, "_value": "\\(2^2\\)"}]}, {"_value": "4"}, {}]}, {"td": [{}, {}, {"_value": "2"}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "25058"}, "_value": "\\(4 \\boldcdot \\frac12=2\\)"}]}]}, {"td": [{}, {}, {"_value": "1"}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "failed-conversion"}, "_value": "\\(2 \u00a0\\boldcdot \\frac12=1\\)"}]}]}, {"td": [{}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "8478"}, "_value": "\\(2^{\\text-1}\\)"}]}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "497"}, "_value": "\\(\\frac{1}{2}\\)"}]}, {}]}, {"td": [{}, {}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "451"}, "_value": "\\(\\frac{1}{4}\\)"}]}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "failed-conversion"}, "_value": "\\(\\frac12 \u00a0\\boldcdot \\frac12 = \\frac14\\)"}]}]}, {"td": [{}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "25113"}, "_value": "\\(2^{\\text-3}\\)"}]}, {}, {}]}, {"td": [{}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "25133"}, "_value": "\\(2^{\\text-4}\\)"}]}, {}, {}]}, {"td": [{}, {}, {"span": [{"_attributes": {"class": ["math", "math-repaired"], "data-png-file-id": "2239"}, "_value": "\\(\\frac{1}{32}\\)"}]}, {}]}]}]}]}
Student Response
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Activity Synthesis
Display the completed table, and resolve any questions students have. Questions for discussion:
“How would you make sense of
\(2^{\text-6}\)
?” (I know
\(2^{\text-5}=\frac{1}{32}\)
, and I know that
\(2^{\text-6}\)
would be the result of multiplying by another
\(\frac12\)
. So,
\(2^{\text-6} = \frac{1}{64}\)
.
“How could we figure out the value of
\(7^{\text-2}\)
?” (Possible response: We could rewrite it as
\(\frac{7^0}{7^2}\)
, and then
\(\frac{1}{49}\)
.


aligns: no