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: Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.
problem_activity: Warm-up
This warm-up highlights the three forms of quadratic expressions students have seen so far. It reinforces how each form gives different insights into the graph of a quadratic function.
Launch
Arrange students in groups of 2. Give students a minute of quiet think time, and then ask them to share their thinking with a partner.
Student Facing
Expressions in different forms can be used to define the same function. Here are three ways to define a function
\(f\)
.
\(f(x)=x^2-4x+3\)
(standard form)
\(f(x)=(x-3)(x-1)\)
(factored form)
\(f(x)=(x-2)^2-1\)
(vertex form)
Which form would you use if you want to find the following features of the graph of
\(f\)
? Be prepared to explain your reasoning.
the
\(x\)
-intercepts
the vertex
the
\(y\)
-intercept
Student Response
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Activity Synthesis
Invite students to share their responses and explanations. Encourage students to name specific features of the graph of
\(f\)
. If time permits, consider asking students to use all of the info they can gather to sketch a graph.
If not already mentioned by students, point out that while the standard form allows us to find the
\(y\)
-intercept more readily, the
\(y\)
-intercept can always be found by evaluating the function at
\(x=0\)
.


aligns: yes