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 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: 
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, assess whether the following problem or activity is 'aligned' with the given Common Core standard in Mathematics.

shot_info: im_task_001477_pos_0

K

fl_problem_001097_pos_0

7

fl_problem_001894_neg_0

HS_0
standard_description: Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
problem_activity: Task
In triangles $ABC$ and $DEF$ below $m(\angle A) = m(\angle D)$, $m(\angle B) = m(\angle E)$ and $|\overline{AB}| = |\overline{DE}|$.
[IMG]
Find a sequence of translations, rotations, and reflections which maps $\triangle ABC$ to $\triangle DEF$.
After working on problem (a), Melissa says
Since $m(\angle A) = m(\angle D)$ and $m(\angle B) = m(\angle E)$ then I also know that $m(\angle C) = m(\angle F)$. So these triangles share all three angles and this is enough to know that they are congruent. I don't need to be told that $|\overline{AB}| = |\overline{DE}|$.
Is Melissa correct that $m(\angle C) = m(\angle F)$? Explain.
Is she right that two triangles sharing three pairs of congruent angles are always congruent? Explain.
Below are two triangles which share three congruent angles: $m(\angle P) = m(\angle T)$, $m(\angle Q) = m(\angle U)$, $m(\angle R) = m(\angle V)$:
[IMG]
Show that, after applying a suitable dilation, $\triangle PQR$ is congruent to $\triangle TUV$.


aligns: no