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


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.

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

shot_info: fl_problem_001894_neg_0

HS_0
standard_description: Solve linear equations in one variable.
problem_activity: Activity
Students represent a scenario with an equation and use the equation to find solutions. They create a graph (either with a table of values or by using two intercepts), interpret points on the graph, and interpret points not on the graph (MP2).
Launch
Allow about 10 minutes quiet think time for questions 1 through 4, then have students work with a partner to discuss questions 4 and 5. Look for students who define the variables or label the axes differently. This can be an opportunity to discuss the importance of defining what quantities your variables represent and that different graphs can represent the same information.
Speaking, Listening: MLR2 Collect and Display.
Listen for and record the language students use to discuss the statement “List two ways that you can tell if a pair of numbers is a solution to an equation.” Organize and group similar strategies in the display for students to refer back to throughout the lesson. For example, group strategies that reference the equation in one area and strategies that reference the graph in another area. This will help students solidify their understanding about solutions to an equation.
Design Principle(s): Support sense-making; Maximize meta-awareness
Student Facing
The Fabulous Fish Market orders tilapia, which costs
$
3 per pound, and salmon, which costs
$
5 per pound. The market budgets
$
210 to spend on this order each day.
What are five different combinations of salmon and tilapia that the market can order?
Define variables and write an equation representing the relationship between the amount of each fish bought and how much the market spends.
Sketch a graph of the relationship. Label your axes.
[IMG]
On your graph, plot and label the combinations A—F.
|                   | A | B | C | D | E | F |
|-------------------|---|---|---|---|---|---|
| pounds of tilapia |   |   |   |   |   |   |
| pounds of salmon  |   |   |   |   |   |   |
Which of these combinations can the market order? Explain or show your reasoning.
List two ways you can tell if a pair of numbers is a solution to an equation.
Student Response
Teachers with a valid work email address can
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for free access to Student Response.
Anticipated Misconceptions
If students let
\(x\)
be pounds of salmon, then the equation would be
\(5x + 3y = 210\)
and the coordinates would be reversed. The intercepts of the graph would be
\((0,70)\)
and
\((42,0)\)
. This is a good place to mention the importance of defining what quantities your variables represent.
Activity Synthesis
Ask students to share some strategies for graphing and features of their graphs. Invite 2–3 students to display their graphs. Consider asking:
"Was the slope of the line positive or negative? Why does that makes sense in this situation?" (In this relationship, as one quantity increases the other must decrease in order to keep the sum
\(3x + 5y\)
constant.)
"What was your strategy for graphing the relationship?" (Plotting points from question 1, using the table, figuring out the intercepts and connecting the line)
"Why does it make sense for the graph to be only in quadrant I?" (You cannot purchase a negative amount of fish, so the
\(x\)
and
\(y\)
values cannot be negative.)
"How is this situation different from the apples and oranges problem in a previous lesson?" (Buying
\(\frac12\)
pound of fish is reasonable while buying
\(\frac12\)
of an apple probably is not. )
Briefly reiterate key concepts: If a point is not on the graph of the equation then it is not a solution. The ordered pairs that are solutions to the equation all make the equation true and are all found on the line that is the graph of the equation.
A discussion could also include the detail that orders that are less than
$
210 can also be considered to work, because there is money left over. That gives an opportunity to discuss the shape of the graph of
\(3x + 5y \le 210\)
.


aligns: no