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: Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point.
problem_activity: Activity
The purpose of this activity is for students to consider the height of points on a circle not centered at
\((0,0)\)
. Working within the context of a Ferris wheel, students use their understanding of the unit circle to identify the height of a point on the wheel relative to the center of the wheel and then combine this value with the given height of the center of the Ferris wheel to calculate the actual height off the ground. This work builds on the previous activity and builds toward transformations of trigonometric functions, which will be used in future lessons.
While approximate answers are acceptable and can be easier to think about than an expression with sine in it, encourage students to write out the exact value for the heights using sine. This practice will help in the future when students write and algebraically transform equations for trigonometric functions.
Digital
Launch
If students have access to devices that can run a GeoGebra applet, suggest that the Unit Circle Slider applet might be a helpful tool in this activity. Consider demonstrating how to use the applet. The slider works to change the angle that the given segment creates with the positive
\(x\)
-axis. The displayed coordinates indicate the coordinates of the intersection point of the line segment with the unit circle.
Display the image of the Ferris wheel for all to see. Have students estimate the height of each position as the Ferris wheel moves counterclockwise. In the activity, they will precisely calculate the value.
\(\frac \pi {12}\)
radians
\(\frac \pi 2\)
radians
\(\frac {5\pi}6\)
radians
\(\frac {5\pi}3\)
radians
Reading: MLR6 Three Reads.
Use this routine to support reading comprehension, without solving, for students. Use the first read to orient students to the situation. Ask students to describe what the situation is about without using numbers (A Ferris wheel with point
\(P\)
on the circle). After the second read, students list any quantities that can be counted or measured, without focusing on specific values (the distance off of the ground, the radius, and the angle of rotation in radians). After the third read, ask students to brainstorm possible strategies to answer the questions. This helps students connect the language in the word problem and the reasoning needed to solve the problem while keeping the cognitive demand of the task.
Design Principle: Support sense-making
Student Facing
The center of a Ferris wheel is 40 feet off the ground, and the radius of the Ferris wheel is 30 feet. Point
\(P\)
is shown at 0 radians.
[IMG]
​​​​​​
Calculate how high off the ground point
\(P\)
is as the Ferris wheel rotates counterclockwise starting from 0 radians.
\(\frac \pi {12}\)
radians
\(\frac \pi 2\)
radians
\(\frac {5\pi}6\)
radians
\(\frac {5\pi}3\)
radians
As you go around on the Ferris wheel, at which position(s) would you be 60 feet off the ground? Explain your reasoning.
You may use this applet if you choose.
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
A Unit Circle Slider applet is available in the digital version of this activity.
Arrange students in groups of 2. Display the image of the Ferris wheel for all to see. Tell them the center of the wheel is 40 feet high and it has a radius of 30 feet. Ask students to give a rough estimate for these questions:
“How high off the ground will point
\(P\)
be after the wheel rotates
\(\frac \pi {12}\)
radians?” (That is a small rotation, so around 50 feet.)
“How high off the ground will point
\(P\)
be after the wheel rotates
\(\frac \pi {2}\)
radians?” (That’s a quarter turn, so
\(P\)
will be at the top, which is a height of 70 feet.)
Give students quiet work time to complete the actual calculations needed for the first question and then time to share their work with a partner before completing the rest of the activity.
Reading: MLR6 Three Reads.
Use this routine to support reading comprehension, without solving, for students. Use the first read to orient students to the situation. Ask students to describe what the situation is about without using numbers (A Ferris wheel with point
\(P\)
on the circle). After the second read, students list any quantities that can be counted or measured, without focusing on specific values (the distance off of the ground, the radius, and the angle of rotation in radians). After the third read, ask students to brainstorm possible strategies to answer the questions. This helps students connect the language in the word problem and the reasoning needed to solve the problem while keeping the cognitive demand of the task.
Design Principle: Support sense-making
Action and Expression: Internalize Executive Functions.
Provide students with a two-column table to organize the given angles of rotation (in radians) and corresponding height off the ground (in feet). The table will also provide a visual support for students to write out exact values using sine.
Supports accessibility for: Language; Organization
Student Facing
The center of a Ferris wheel is 40 feet off of the ground, and the radius of the Ferris wheel is 30 feet. Point
\(P\)
is shown at 0 radians.
[IMG]
Calculate how high off the ground point
\(P\)
is as the Ferris wheel rotates counterclockwise starting from 0 radians.
\(\frac \pi {12}\)
radians
\(\frac \pi 2\)
radians
\(\frac {5\pi}6\)
radians
\(\frac {5\pi}3\)
radians
As
\(P\)
goes around on the Ferris wheel, estimate which angle(s) of rotation put
\(P\)
60 feet off the ground. Explain your reasoning.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Anticipated Misconceptions
For students having difficulty figuring out different rotations or heights on the Ferris wheel, suggest that they add more labels to the diagram, such as some of the known heights.
Activity Synthesis
Begin the discussion by asking, “Would you ever be 75 feet off the ground? Explain your reasoning.” (No, the highest you would ever be is 70 feet, which is directly above the center of the Ferris wheel rotated
\(\frac \pi 2\)
radians from 0 radians.)
Invite students to share how they estimated the angles of rotation that put
\(P\)
60 feet off the ground. Highlight students who:
labeled the diagram of the Ferris wheel in order to identify that there are 2 angles, one in quadrant 1 and one in quadrant 2
used the unit circle display to identify angles where the value of sine is close to
\(\frac{2}{3}\)
, since 60 feet is 20 feet higher than the center and 20 feet is
\(\frac{2}{3}\)
the radius of the wheel
wrote an equation to represent the situation, such as
\(60=40 + 30 \sin(\theta)\)
For any of these methods that were not used, ask students to consider them now. A key point here is recognizing that the angle needed is one where
\(\frac23\)
the length of the radius is added to the height of the center of the wheel.


aligns: no