few_shots: 
Example 1:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Description: 
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.

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


Description: 
Solve quadratic equations in one variable.

Answer: 
Yes. This task shows one quadratic equation with one variable, x.

Example 3:
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 $P$.


Description: 
Construct a tangent line from a point outside a given circle to the circle.

Answer: 
No. This task does not involve drawing lines that are tangent to circles. The only line that may be drawn is the ray between points P and B.

Now, assess whether the following problem or activity aligns with the provided description.

shot_info: fl_problem_001894_neg_0

HS_0

im_task_000870_pos_0

HS_1

im_task_001160_neg_0

HS_2
standard_description: Apply properties of operations as strategies to multiply and divide. (Students need not use formal terms for these properties.) ^( )Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
problem_activity: Problem 1
How many?
[IMG]
Problem 2
Determine the value of the unknown in each equation below.
a.
$$a = 7 \times 8$$
b.
$$42 \div b =7$$
c.
$$49 = 7 \times c$$
d.
$$d\div 7 = 9$$
Problem 3
Here is the partially filled-in multiplication table from Lessons 1 and 6.
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text-align:center"}, "_value": "15"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "20"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "25"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "30"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "35"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "40"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "45"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "50"}]}, {"td": [{"_attributes": {"style": "background-color: rgb(153, 204, 255); text-align: center;"}, "strong": [{"_value": "6"}]}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "6"}, {"_attributes": {"style": "background-color:#ffffff; text-align:center"}, "_value": "12"}, {"_attributes": {"style": "background-color:#ffffff; 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Add the new facts you encountered in this lesson to your multiplication table.


aligns: no