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.


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


Description: 
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:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Description: 
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, assess whether the following problem or activity aligns with the provided description.

shot_info: im_task_001477_neg_0

K

fl_problem_001097_neg_0

7

fl_problem_001894_pos_0

HS_0
standard_description: Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.
problem_activity: Activity
This task prompts students to carefully interpret a description and some expressions representing exponential decay. It further reinforces the distinction between how exponential and linear functions change over intervals of time, which was explored in the previous activity.
The decay factor is not given for a familiar unit of input (that is, the caffeine decays by half every 6 hours, instead of every hour). To correctly represent the decay with functions, students need to notice that the
\(\frac12\)
does not correspond to the change in 1 hour, and that the decay factor in 1 hour is not
\(\frac16\)
of
\(\frac12\)
.
Launch
Ask students to close their books or devices. Tell the class they are going to do some calculations around how the amount of caffeine in the body changes over time after consuming a certain amount. Before students read the task statement or do any calculations, ask “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Poll the class for their estimates and display the result for all to see. Students will consider this question again during the whole-class discussion.
Tell students to open their books or devices and read the opening statement of the activity. Explain that the term
half-life
refers to the time it takes for a certain quantity (such as the concentration of a substance in a body) that is decaying exponentially to decrease by half. In this context, a half-life of 6 hours means that the amount of caffeine in the body is cut in half 6 hours after it is consumed.
Pause students after they complete the first set of questions. Discuss what the 100, the
\(\frac12\)
, and the exponent in each expression represent, and review students’ responses to the questions before they continue with the task.
Action and Expression: Internalize Executive Functions.
Provide students with a three-column graphic organizer that leaves space at the top for general notes. Label each column with the headers of “half-life,” “hours,” and “expression.”  Encourage students to write important information in the top space, such as the starting amount and the ratio of hours per half-life. Demonstrate organizing the first question into the columns, placing the number of half-lives (1) in the first column, the number of hours (6) in the second, and the expression (
\(100 \boldcdot (\frac{1}{2})^1\)
) in the third. When students arrive at the last question, ask them what is the unit that they will be using as the base for their function (the hour) and what information on their organizer can help them to derive their function (the expression for one hour). Then, refer back to the question and ask them what they would have to do to write it as a function for
\(h\)
hours.
Supports accessibility for: Language; Organization
Student Facing
In healthy adults, caffeine has an average half-life of about 6 hours. Let's suppose a healthy man consumes a cup of coffee that contains 100 mg of caffeine at noon.
Each of the following expressions describes the amount of caffeine in the man's body some number of hours after consumption. How many hours after consumption?
\(100 \boldcdot \left(\frac12\right)^1\)
\(100 \boldcdot \left(\frac12\right)^3\)
\(100 \boldcdot \left(\frac12\right)^{\frac16}\)
\(100 \boldcdot \left(\frac12\right)^t\)
Write a function
\(g\)
to represent the amount of caffeine left in the body,
\(h\)
hours after it enters the bloodstream.
The function
\(f\)
represents the amount of caffeine left in the body after
\(t\)
6-hour periods. Explain why
\(g(6) = f(1)\)
.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
What percentage of the initial amount of caffeine do you expect to break down in the first 3 hours: less than 25%, exactly 25%, or more than 25%? Explain or show your reasoning.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
Students may struggle viewing groups of 6 hours as a unit for the half-life of caffeine. Identifying that
\(100 \boldcdot \left(\frac{1}{2}\right)^{\frac{1}{6}}\)
is the number of mg of caffeine left after 1 hour is the critical step. If students do not see what this expression means in the situation, ask them how any times the scale factor
\(\left(\frac{1}{2}\right)^{\frac{1}{6}}\)
has to be applied for the caffeine to be cut in half. Then ask them how many hours it takes for the caffeine to be cut in half.
Activity Synthesis
Begin the discussion by repeating the earlier question, “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Invite students to share their reasoning, making sure they see that “24 hours later” is 4 periods of 6 hours, so the amount of caffeine will have been halved 4 times:
\(100 \boldcdot \left(\frac12\right)^4 = 100 \boldcdot \frac{1}{16} = 6.25\)
.
Next, focus the discussion on the distinction between expressing the amount of caffeine as a function of time in groups of 6 hours (function
\(f\)
) and time in groups of 1 hour (function
\(g\)
). Discuss with students:
“What decay factor did you use to write the expression for function
\(g\)
? What is the approximate decay rate per hour?” (Raise
\(\frac12\)
to an exponent of
\(\frac 16\)
, which is approximately 0.89, so the decay rate is about 11%.)
“Why not divide
\(\frac12\)
by 6 (or multiply it by
\(\frac16\)
) to get the decay factor for an hour?” (Because when that factor is raised to an exponent of 6, it needs to equal
\(\frac12\)
.
\(\left(\frac{1}{12}\right)^6\)
is not
\(\frac12\)
.)
“How can we show that the expression for
\(g\)
, which is
\(100 \boldcdot \left(\left(\frac12\right)^{\frac16}\right)^h\)
, is equivalent to the expression for
\(f\)
, which is
\(100 \boldcdot \left(\frac12\right)^t\)
?” (We know that
\(h=6t\)
. If we substitute
\(6t\)
in the expression for
\(g\)
, we have
\(100 \boldcdot \left(\frac12\right)^{(\frac16)^{6t}}\)
, which equals
\(100 \boldcdot \left(\frac12\right)^t\)
.)
Time permitting, consider showing the two graphs representing the two functions.
Function
\(g\)
[IMG]
Function
\(f\)
[IMG]
Here are some possible questions for discussion about the graphs:
“How are these two graphs alike and different?” (They seem to be the same curve. They have the same vertical intercept. The scales for the horizontal axis are different. One curve goes through
\((1,50)\)
and the other goes through
\((6,50)\)
.)
“On the graph representing
\(f\)
, what does the point at
\((1,50)\)
mean in the context? What is the corresponding point on the graph of
\(g\)
?” (The amount of caffeine in the body after 1 period of 6 hours is 50 milligrams. This is the same as the point
\((6,50)\)
on the graph of
\(g\)
.)
“How could these two graphs represent the same decay, if the coordinates they go through are not the same?” (They use different units for the input. 1 unit of the input in
\(f\)
is 6 units of the input in
\(g\)
.)
“To find the amount of caffeine 3 hours after consumption, what input value should we use for each function?” (
\(\frac12\)
for
\(f\)
and 3 for
\(g\)
) “What about for 30 minutes after consumption?” (
\(\frac {1}{12}\)
for
\(f\)
and
\(\frac12\)
for
\(g\)
)
Reading, Writing, Speaking: MLR3 Clarify, Critique, Correct.
Before students share their responses for the function
\(g\)
, present an incorrect answer and explanation. For example, “We know the amount of caffeine in the body is cut in half 6 hours after it is consumed, which is represented by the function
\(f(t)=100\boldcdot(\frac{1}{2})^t\)
. To find the decay factor per hour, we can divide
\(\frac{1}{2}\)
by 6 to get
\(\frac{1}{12}\)
. This means that the amount of caffeine left in the body,
\(h\)
hours after it enters the bloodstream, can be represented by the function
\(g(h)=100\boldcdot(\frac{1}{12})^h\)
.” Ask students to identify the error, critique the reasoning, and write a correct explanation. As students discuss with a partner, listen for students who clarify the meaning of the decay factor per hour. Invite students to share their critiques and corrected explanations with the class. This will help students evaluate, and improve upon, the written mathematical arguments of others.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness


aligns: yes