few_shots: 
Example:
Problem/activity:
Narrative
The purpose of this activity is for students to create composite shapes and compose new shapes from composite shapes. For example, students may find that a rectangular prism can be made from four triangular prisms or from eight smaller cubes. Students then discover that they can combine two of the composed rectangular prisms to make a new rectangular prism.
Required Materials
Materials to Gather
Geoblocks
Launch
Groups of 4
Give each group geoblocks.
Display a large cube.
“This cube can be built from other shapes. See how many ways you can build a cube using other shapes.”
3 minutes: partner work time
Share responses.
Activity
“Choose a different shape. Try to find smaller shapes you can use to build that shape.”
5 minutes: partner work time
Student Response
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for free access to Student Response.
Activity Synthesis
Display a cube and a composite cube—for example, a cube composed of four small cubes and one rectangular prism.
Demonstrate putting the shapes together.
“What new shape did I make with the cube and the cube I made?”
“Put your shapes together to make a rectangular prism.”
Display a rectangular prism and two triangular prisms.
“Find these shapes. Put them together. What new shape can you make?” (a cube, a rectangular block)


Description: 
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape. (Students do not need to learn formal names such as “right rectangular prism.”)

Answer: 
Yes. This activity asks students to compose three-dimensional shapes, using geoblocks, to create composite shapes.

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

shot_info: im_lesson_009431_pos_0

1
standard_description: Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
problem_activity: Activity
By now, students recognize that when a quadratic equation is in the form of
\(\text{expression} = 0\)
and the expression is in factored form, the equation can be solved using the zero product property. In this activity, they encounter equations in which one side of the equal sign is not 0. To make one side equal 0 requires rearrangement. For example, to solve
\(x(x+6)=8\)
, the equation needs to be rearranged to
\(x(x+6)-8=0\)
. Yet because the expression on the left is no longer in factored form, the zero product property won’t help after all and another strategy is needed.
Students recall that to solve a quadratic equation in the form of
\(\text{expression} = 0\)
is essentially to find the zeros of a quadratic function defined by that expression, and that the zeros of a function correspond to the horizontal intercepts of its graph. In the case of
\(x(x+6)-8=0\)
, the function whose zeros we want to find is defined by
\(x(x+6)-8\)
. Graphing
\(y=x(x+6)-8\)
and examining the
\(x\)
-intercepts of the graph allow us to see the number of solutions and what they are.
Launch
Arrange students in groups of 2 and provide access to graphing technology. Give students a moment to think quietly about the first question and then ask them to briefly discuss their response with their partner before continuing with the rest of the activity.
Action and Expression: Provide Access for Physical Action.
Support effective and efficient use of tools and assistive technologies. To use graphing technology, some students may benefit from a demonstration or access to step-by-step instructions.
Supports accessibility for: Organization; Memory; Attention
Student Facing
Han is solving three equations by graphing.
\((x-5)(x-3)=0\)
\(\displaystyle (x-5)(x-3)=\text-1\)
\(\displaystyle (x-5)(x-3)=\text-4\)
To solve the first equation,
\((x-5)(x-3)=0\)
, he graphed
\(y=(x-5)(x-3)\)
and then looked for the
\(x\)
-intercepts of the graph.
Explain why the
\(x\)
-intercepts can be used to solve
\((x-5)(x-3)=0\)
.
What are the solutions?
To solve the second equation, Han rewrote it as
\((x-5)(x-3)+1=0\)
. He then graphed
\(y=(x-5)(x-3)+1\)
.
Use graphing technology to graph
\(y=(x-5)(x-3)+1\)
. Then, use the graph to solve the equation. Be prepared to explain how you use the graph for solving.
Solve the third equation using Han’s strategy.
Think about the strategy you used and the solutions you found.
Why might it be helpful to rearrange each equation to equal 0 on one side and then graph the expression on the non-zero side?
How many solutions does each of the three equations have?
Student Response
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for free access to Student Response.
Student Facing
Are you ready for more?
The equations
\((x-3)(x-5)=\text-1\)
,
\((x-3)(x-5)=0\)
, and
\((x-3)(x-5)=3\)
all have whole-number solutions.
Use graphing technology to graph each of the following pairs of equations on the same coordinate plane. Analyze the graphs and explain how each pair helps to solve the related equation.
\(y=(x-3)(x-5)\)
and
\(y=\text-1\)
\(y=(x-3)(x-5)\)
and
\(y=0\)
\(y=(x-3)(x-5)\)
and
\(y=3\)
Use the graphs to help you find a few other equations of the form
\((x-3)(x-5)=z\)
that have whole-number solutions.
Find a pattern in the values of
\(z\)
that give whole-number solutions.
Without solving, determine if
\((x-5)(x-3)=120\)
and
\((x-5)(x-3)=399\)
have whole-number solutions. Explain your reasoning.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
If students enter the equation
\((x-5)(x-3)=0\)
into their graphing technology, they may see an error message, or they may see vertical lines. The lines will intersect the
\(x\)
-axis at the solutions, but they are clearly not graphs of a quadratic function. Emphasize that we want to graph the function defined by
\(y=(x-5)(x-3)\)
and use its
\(x\)
-intercepts to find the solution to the related equation.
All
the points on the two vertical lines do represent solutions to the equation (because the points along each vertical line satisfy the equation regardless of the value chosen for
\(y\)
), but understanding this is beyond the expectations for students in this course.
Activity Synthesis
Invite students to share their responses, graphs, and explanations on how they used the graphs to solve the equations. Discuss questions such as:
“
Are the original equation
\((x-5)(x-3)=\text-1\)
and the rewritten one
\((x-5)(x-3)+1=0\)
equivalent?” (Yes, each pair of equations are equivalent. In that example, 1 is added to both sides of the original equation.)
"Why might it be helpful to rearrange the equation so that one side is an expression and the other side is 0?" (It allows us to find the zeros of the function defined by that expression. The zeros correspond to the
\(x\)
-intercepts of the graph.)
“What equation would you graph to solve this equation:
\((x-4)(x-6)=15\)
?” (
\(y=(x-4)(x-6)-15\)
) “What about
\((x+3)^2-1=5\)
?" (
\(y=(x+3)^2-6\)
)
Make sure students understand that some quadratic functions have two zeros, some have one zero, and some have no zeros, so their respective graphs will have two, one, or no horizontal intercepts, respectively.
Likewise, some quadratic equations have two solutions, some have one solution, and some have no real solutions. (Because students won’t know about numbers that aren’t real until a future course, for now it is sufficient to say “no solutions.”)


aligns: no