few_shots: 
Example 1:
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.

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.


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

Example 2:
Concept/skill: 
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

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
$$-\frac{1}{4}$$


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:
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.

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


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_neg_0

K

fl_problem_001097_pos_0

7

fl_problem_001894_neg_0

HS_0
standard_description: Interpret expressions that represent a quantity in terms of its context.
problem_activity: Activity
The mathematical purpose for this activity is to allow students to practice solving for a variable in an equation with two variables. In the associated Algebra lesson, expressing one variable in terms of another is necessary before substituting an expression for a variable in another equation.
Launch
Demonstrate how to solve for a variable by showing students the equation
\(a + 2b = 5\)
. Solving for
\(a\)
results in
\(a = 5 - 2b\)
or
\(a = \text{-}2b + 5\)
. Solving for
\(b\)
results in
\(b = \frac{5-a}{2}\)
or
\(b = \frac{5}{2} - \frac{a}{2}\)
or another equivalent expression.
Student Facing
Solve for the indicated variable.
Solve for
\(k\)
.
\(2t+k=6\)
Solve for
\(n\)
.
\(10n=2p\)
Solve for
\(c\)
.
\(12-6d=3c\)
Solve for
\(g\)
.
\(h=8g+4\)
Solve for
\(x\)
.
\(4x+3y=12\)
Solve for
\(y\)
.
\(4x+3y=12\)
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Activity Synthesis
The purpose of the discussion is to share strategies for solving for one variable in an equation including more than one variable. Select students to share their answers and strategies. As students share, consider asking another student why each step makes sense to do.
Some questions for discussion:
"How do you know when you are done solving for a variable?" (When the variable I want to solve for is the only thing on one side of the equation.)
"How is solving for
\(x\)
in an equation like
\(4x+3y=12\)
similar or different from solving for
\(x\)
in an equation like
\(4x + 6 = 12\)
?" (The steps are very similar, but when there is only one variable, we can know the value of
\(x\)
that makes the equation true.)
"If you know the value for
\(y\)
is 2 and need to know the value for
\(x\)
, would you rather begin with the equation
\(4x + 3y = 12\)
or the equivalent equation
\(x = 3 - \frac{3}{4}y\)
? Explain your reasoning." (I would rather use the second equation since I only need to substitute in the value for
\(y\)
and evaluate the expression rather than solve an equation.)


aligns: no