few_shots: 
Example 1:
Problem/activity:
You’re mixing ingredients for cookies. The recipe says to combine 6 tablespoons, or ¾ stick, of butter with 1 cup of sugar. You accidentally mix in a whole stick of butter (8 tablespoons) with the cup of sugar. How can you fix this?


Concept/skill: 
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.

Answer: 
Yes. This problem solves a real-world problem involving mixing ingredients for cookies. A student doing this problem would need to reason about equivalent ratios of butter amounts to sugar amounts.

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:
Problem 1
The table below contains single equations in a single variable, single equations in two variables, and pairs of simultaneous equations in two variables. Check the appropriate box to show whether there is no solution, exactly 1 solution, or if there is an infinite number of solutions.
{"table": [{"_attributes": {"align": "center", "class": ["table", "table-bordered"], "style": "table-layout:fixed; width:70%"}, "tbody": [{"tr": [{"td": [{"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "No solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Exactly 1 solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Infinite number of solutions"}]}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9=4x+7}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y=3y+5}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y={x\\over2}}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${3x+6y-9=0}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${x=y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${1.25x=1.25y}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9.9=6x+8y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9x=2.5y-8.8}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=1.5}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=2.5}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {}]}]}]}]}
Problem 2
Kim has a small container and a large container as shown.
[IMG]
It takes 16 of the small containers to fill the large container. Three small containers leave 1.95 gallons of space in the big container.
What is the size of each of the two containers?


Concept/skill: 
Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.

Answer: 
No. Completing neither of these problems requires replacing one equation by the sum of that equation and a multiple of the other equation.

Now, does the problem or activity shown below enable students to learn the full intent of the following concept or skill?

shot_info: fl_problem_000972_pos_0

6

fl_problem_001097_neg_0

7

fl_problem_001392_neg_0

8
standard_description: For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
problem_activity: Activity
The goal of this activity is to write an expression that models the volume of a box as a function of the side length
\(x\)
of the square cutouts. Students then graph the function, use the graph to estimate the value of
\(x\)
that will produce the box with the largest volume, and establish a reasonable domain for the function by considering the context.
During the discussion, students interpret their expression in context (MP2). Of particular interest, values of
\(x\)
greater than 5.5 inches result in positive volumes even though cutouts of this size do not make sense when constructing a box from an 8.5 inch by 11 inch sheet of paper. There is no need for students to rewrite the expression for
\(V(x)\)
at this time in standard form. In later lessons, students will practice rewriting equations for polynomials in different forms.
Launch
Arrange students in groups of 2. Tell students there are many possible answers for the first question. After quiet work time, ask students to compare their response to their partner’s and decide which plan they like best. Select 2–3 groups to share their plans. Provide access to devices that can run Desmos or other graphing technology.
Representation: Internalize Comprehension.
Use annotations to highlight connections between representations in a problem. For example, prior to developing an equation for volume, have students label the edges of their box with the calculations they used to determine the length, width, and height.
Supports accessibility for: Visual-spatial processing
Student Facing
[IMG]
The volume
\(V(x)\)
in cubic inches of the open-top box is a function of the side length
\(x\)
in inches of the square cutouts. Make a plan to figure out how to construct the box with the largest volume.
Pause here so your teacher can review your plan.
Write an expression for
\(V(x)\)
.
Use graphing technology to create a graph representing
\(V(x)\)
. Approximate the value of
\(x\)
that would allow you to construct an open-top box with the largest volume possible from one piece of paper.
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?
The surface area
\(A(x)\)
in square inches of the open-top box is also a function of the side length
\(x\)
in inches of the square cutouts.
Find one expression for
\(A(x)\)
by summing the area of the five faces of our open-top box.
Find another expression for
\(A(x)\)
by subtracting the area of the cutouts from the area of the paper.
Show algebraically that these two expressions are equivalent.
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
Some students may forget that the width and length of the paper are based on removing length
\(x\)
twice due to cutting out squares at each corner. Encourage these students to revisit their work from the previous activity and think about how they can calculate length and width without measuring directly.
If students have trouble finding a general expression for the volume of the box, encourage them to revisit their results from the previous activity and write the volumes of the boxes without simplifying their answers (e.g., 6.5 in
\(\times\)
9 in
\(\times\)
1 in instead of 58.5 in
3
) in order to see the pattern more easily.
Activity Synthesis
The purpose of this discussion is for students to think critically about the equation they’ve written to model the volume of the box. Here are some questions for discussion:
“What is a value of
\(x\)
that wouldn’t make sense?” (
\(x=6\)
would not make sense, because the sheet of paper is only 8.5 inches long on one side, so cutting out 6 inch by 6 inch squares from each corner is not possible.)
“What is the smallest value that would work for
\(x\)
? The largest?” (The smallest value is some number greater than 0 inches that you could still use as a side length to cut out the squares and fold up the box sides to make a very short box. The largest value is some number less than half of 8.5 inches that you could still use as a side length to cut out the squares and fold up the box sides to make a very tall box.)
“What is the range of
\(x\)
values where the volume is increasing? Decreasing?” (The volume increases as the cutouts get larger, until
\(x\)
is about 1.6 inches. Then the volume decreases until
\(x\)
is 4.25 inches, which creates a box with a volume of 0. The volume also starts increasing again when
\(x\)
is about 4.9 inches, but these values of
\(x\)
don’t make sense because we can’t make cutouts that large.)
If no students bring up the word
domain
during the discussion, remind students of the meaning of the word and select students to recommend a reasonable domain for the function
\(V\)
.
Conclude the discussion by telling students that
\(V\)
is a type of
polynomial
function and that they will learn about other situations polynomial functions can model along with what graphs of polynomial functions can look like in future lessons.
Conversing: MLR2 Collect and Display.
During the discussion, listen for and collect the language students use to interpret the equation they’ve written to model the volume of the box. Call students’ attention to language such as “range of values,” “groups of parentheses,” or other key parts of the equation representing the length, width, or height. Write the students’ words and phrases on a visual display and update it throughout the remainder of the lesson. Use this opportunity to define a “polynomial function.” Remind students to borrow language from the display as needed. This will help students use mathematical language to interpret the equation for the volume of the box.
Design Principle(s): Maximize meta-awareness


aligns: yes