few_shots: 
Example 1:
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.

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


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.

Example 2:
Concept/skill: 
Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

Problem/activity:
Task
Solve the quadratic equation
$$ x^2 = (2x - 9)^2 $$
using as many different methods as possible.


Answer: 
No. This task does not involve an equation in two variables. There is only one variable, x.

Example 3:
Concept/skill: 
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Problem/activity:
Task
Suppose $A$ is an angle with vertex $P$, as pictured below:
[IMG]
Draw a circle with center $P$ and with  radius $r \gt 0$.  Explain why the circle
meets each ray of angle $A$ in a single point.  Label these points $Q$ and $R$ respectively.
Draw circles with centers $Q$ and $R$ respectively and radius $r$.  These circles meet at $P$ and a second point to be labelled $B$.
Show that ray $\overrightarrow{PB}$  bisects angle $P$.


Answer: 
Yes. This task asks students to draw a ray that bisects an angle. The act of drawing circles and rays may involve compass and straightedge or dynamic geometric software.

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

shot_info: fl_problem_001894_neg_0

HS_0

im_task_000870_neg_0

HS_1

im_task_001160_pos_0

HS_2
standard_description: Interpret multiplication as scaling (resizing), by:
- Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n×a)/(n×b) to the effect of multiplying a/b by 1.
- Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
problem_activity: Narrative
The purpose of this activity is for students to find the missing value in equations involving a whole number and a fraction. They find missing factors or products to make equations true. Many of the problems encourage students to think in steps, interpreting a fraction
\(\frac{a}{b}\)
in terms of division by the denominator and multiplication by the numerator.
Representation: Internalize Comprehension
. Synthesis: Invite students to identify which details were most important to solve the problems. Display the sentence frame: “The next time I solve multiplication problems of a whole number by a fraction, I will pay attention to . . . .“
Supports accessibility for: Conceptual Processing, Memory
Launch
Groups of 2
Activity
5–8 minutes: independent work time
1–2 minutes: partner discussion
Monitor for:
partners who use different strategies.
students who are challenged but persevere and arrive at a viable solution.
Student Facing
Fill in the blanks to make each equation true. Be prepared to explain your reasoning.
\(\frac{1}{3} \times 18 =\underline{\hspace{0.7cm}}\)
\(\frac{7}{9} \times \underline{\hspace{0.7cm}} = \frac{21}{9}\)
\(\frac{1}{15} \times \underline{\hspace{0.7cm}} = 2\)
\(9 \times  6 \frac{2}{3}  = \underline{\hspace{0.7cm}} \)
\(14 \frac {99}{100} \times 10 = \underline{\hspace{0.7cm}} \)
\(7 \frac {3}{5} \times 6 = \underline{\hspace{0.7cm}}\)
\(4 \times 6 \frac {9}{10} = \underline{\hspace{0.7cm}}\)
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Advancing Student Thinking
If students do not get the correct solution for the missing factor problems, ask “How would you represent this expression with a diagram?”
Activity Synthesis
Ask previously selected students to share their responses.
“What was different about you and your partner's strategies?”
“Which problem did you find most challenging and why?”
“Which problem made the most sense to you and why?”


aligns: yes