few_shots: 
Example 1:
Concept/skill: 
Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.

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


Answer: 
Yes. An initial step when solving for x in this problem requires the use of the addition formula for cosine on the left side of the equation.

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_pos_0

HS_0

im_task_000870_neg_0

HS_1

im_task_001160_pos_0

HS_2
standard_description: Recognize volume as an attribute of solid figures and understand concepts of volume measurement.
problem_activity: Narrative
The purpose of this activity is for students to practice multiplying multi-digit numbers that have one or more digits of 0 at the end. Monitor for students who:
use the standard algorithm to evaluate
\(6,\!700 \times 89\)
.
multiply the product
\(67 \times 89\)
by 10 to find the value of the product
\(670 \times 89\)
.
multiply the product of
\(670 \times 89\)
by 10 to find the value of the product
\(6,\!700 \times 89\)
.
Students who observe that
\(670 = 10 \times 67\)
and
\(6,\!700 = 10 \times 670\)
and use these relationships to find the values of the products are observing regularity in repeated reasoning and using their knowledge of how to multiply a whole number by 10 (MP7, MP8).
Representation: Internalize Comprehension.
Synthesis: Invite students to identify which details were important to find products of multi-digit numbers that have zero digits. Display the sentence frame, “The next time I multiply a number that contains zero digits, I will pay attention to . . . .“
Supports accessibility for: Conceptual Processing, Memory
Launch
Groups of 2
Activity
5–7 minutes: independent work time
5–7 minutes: partner discussion
Student Facing
Find the value of each product.
1.
[IMG]
2.
[IMG]
3.
[IMG]
4.
[IMG]
Student Response
Teachers with a valid work email address can
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for free access to Student Response.
Activity Synthesis
Display the product
\(6,\!700 \times 89\)
:
Ask previously identified students to share their solutions.
Display student work or the image from the sample response:
[IMG]
“What is the relationship between
\(670\times89\)
and
\(6,\!700\times89\)
?” (The product
\(6,\!700\times89\)
is ten times larger because one of the factors is ten times greater.)
“What is the relationship between
\(67 \times 89\)
and
\(6,\!700 \times 89\)
?” (The product
\(6,\!700 \times 89\)
is 100 times as large as
\(67 \times 89\)
since 6,700 is
\(100 \times 67\)
.)


aligns: no