few_shots: 
Example 1:
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.

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?


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:
Concept/skill: 
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.

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: 
No. The expressions shown in this problem do not involve multiplying fractions with other fractions, nor whole numbers with fractions.

Example 3:
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.

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?


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

shot_info: fl_problem_000972_pos_0

6

fl_problem_001097_neg_0

7

fl_problem_001392_neg_0

8
standard_description: Derive the formula A = 1/2 ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.
problem_activity: Warm-up
In this warm-up, students compare and contrast two ways of decomposing and rearranging a parallelogram on a grid such that its area can be found. It serves a few purposes: to reinforce the work done in the previous lesson; to allow students to practice communicating their observations; and to shed light on the features of a parallelogram that are useful for finding area—a base and a corresponding height.
The flow of key ideas—to be uncovered during discussion and gradually throughout the lesson—is as follows:
There are multiple ways to decompose a parallelogram (with one cut) and rearrange it into a rectangle whose area we can determine.
The cut can be made in different places, but to compose a rectangle, the cut has to be at a
right angle
to two opposite sides of the parallelogram.
The length of one side of this newly composed rectangle is the same as the length of one side of the parallelogram. We use the term
base
to refer to this side.
The length of the other side of the rectangle is the length of the cut we made to the parallelogram. We call this segment a
height
that corresponds to the chosen base.
We use these two lengths to determine the area of the rectangle, and thus also the area of the parallelogram.
As students work and discuss, identify those who make comparisons in terms of the first two points so they could share later. Be sure to leave enough time to discuss the first four points as a class.
Digital
Launch
Arrange students in groups of 2. Give students 2 minutes of quiet think time and access to geometry toolkits. Ask them to share their responses with a partner afterwards.
If using the digital activity, have students explore the applet for 3 minutes individually, and then discuss with a partner. Students will be able to see the cuts by dragging the point on the segment under the parallelogram.
Student Facing
Elena and Tyler were finding the area of this parallelogram:
[IMG]
Move the slider to see how Tyler did it:
Move the slider to see how Elena did it:
How are the two strategies for finding the area of a parallelogram the same? How they are different?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Print
Launch
Arrange students in groups of 2. Give students 2 minutes of quiet think time and access to geometry toolkits. Ask them to share their responses with a partner afterwards.
If using the digital activity, have students explore the applet for 3 minutes individually, and then discuss with a partner. Students will be able to see the cuts by dragging the point on the segment under the parallelogram.
Student Facing
Elena and Tyler were finding the area of this parallelogram:
[IMG]
Here is how Elena did it:
[IMG]
Here is how Tyler did it:
[IMG]
How are the two strategies for finding the area of a parallelogram the same? How they are different?
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
Ask a few students to share what was the same and what was different about the methods they observed.
Highlight the following points on how Elena and Tyler's approaches are the same, though do not expect students to use the language. Instead, rely on pointing and gesturing to make clear what is meant. If any of these are not mentioned by the students, share them.
The rectangles are identical; they have the same side lengths. (Label the side lengths of the rectangles.)
The cuts were made in different places, but the length of the cuts was the same. (Label the lengths along the vertical cuts.)
The horizontal sides of the parallelogram have the same length as the horizontal sides of the rectangle. (Point out how both segments have the same length.)
The length of each cut is also the distance between the two horizontal sides of the parallelogram. It is also the vertical side length of the rectangle. (Point out how that distance stays the same across the horizontal length of the parallelogram.)
[IMG]
Begin to connect the observations to the terms
base
and
height
. For example, explain:
“The two measurements that we see here have special names. The length of one side of the parallelogram—which is also the length of one side of the rectangle—is called a
base
. The length of the vertical cut segment—which is also the length of the vertical side of the rectangle—is called a
height
that corresponds to that base.”
“Here, the side of the parallelogram that is 7 units long is also called a base. In other words, the word
base
is used for both the segment and the measurement.”
Tell students that we will explore bases and heights of a parallelogram in this lesson.


aligns: no