Del Oro Mathematics

Courses  Pre Algebra  Algebra 1/Math A  Geometry  Preparatory Algebra 2  Algebra 2  EAP Math  Trig/Pre Calc  AP Calculus (AB and BC)  AP Statistics

Curriculum Path Alg 1/Math A PreAlg GeometryAlgebra 2AP StatsTrig/Pre-Calc AP Calc AB/BC Algebra 1 Challenge Test Prep Alg 2EAP Math

Schedule Choice

Math 1  Algebra 1 Credit (75% or better) – 90% - 100% = A – 80% - 89% = B – 75% - 79% = C  Math A Credit – 66% - 74% = A – 57% - 65% = B – 50% - 56% = C

Geometry and UC The decision to include the Geometry requirement was based on recommendations from a Task Force convened in 2007 to study the UC/CSU areas “c” and “d” (mathematics and laboratory science) requirements. This task force included faculty from UC, CSU, California Community Colleges, and California high schools. It agreed that Geometry is integral to college preparatory mathematics education, and that the courses that follow a high school Geometry course typically do not contain much of the critical material and therefore are not considered substitutes for Geometry. In other words, they do not “validate” Geometry. The important topics in Geometry for college readiness are discussed in the 2010 Intersegmental Committee of Academic Senate’sStatement of Competencies in Mathematics Expected of Entering College Students.Statement of Competencies in Mathematics Expected of Entering College Students

Realities of Common Core 1.Common core requires a much deeper understanding of mathematics that students use. 2.The problems are “real world” and not just a collection of algorithms. 3.60% of the assessment is “Performance Based”

4 Claims  Claim 1: Concepts and Procedures – Students can explain and apply mathematical concepts and interpret and carry out mathematical procedures with precision and fluency.  Claim 2: Problem Solving – Students can solve a range of complex, well-posed problems in pure and applied mathematics, making productive use of knowledge and problem-solving strategies.  Claim 3: Communicating Reason – Students can clearly and precisely construct viable arguments to support their own reasoning and to critique the reasoning of others.  Claim 4: Modeling and Data Analysis – Students can analyze complex, real-world scenarios and can construct and use mathematical models to interpret and solve problems.

What Changes? Typical Problem from current State Geometry assessments: 6 in A right circular cone has a height of 6 inches and a diameter of 6 inches. What is the volume of the cone? (A) (B) (C) (D)

A Common Core Type Question An ice cream parlor sells ice cream cones which have three flavors of ice cream. The bottom of the cone is filled with vanilla ice cream so that half of the volume of the cone is vanilla. The top part of the cone is filled with strawberry ice cream. Finally to top the cone off a hemisphere (half of a sphere) of chocolate ice cream is placed on top to the cone. The cone has a height (h) of 6 inches and a diameter (d) of 6 inches. (a) What is the volume of all the ice cream used for the cone? (b) Explain why the height of the vanilla in the cone is not 3 inches. h d T (c) The company that owns the ice cream parlor makes several versions of the ice cream treat which has similar dimensions (the diameter and the height of the cone are equal). To simplify ordering they want to find a formula that will find the volume of the ice cream (both the cone and the hemisphere) based on the total height (T) of the ice cream treat. Find a formula that will calculate the volume based on the value of T.

Answers to Problems  Star Test Type – Answer B  Common Core Problem a) The volume = 36  in 3 (18  for the cone and 18  for the hemisphere) b) The cone is much narrower on the bottom than the top. If we fill the vanilla to only 3 inches, the volume would be which is not 9  in 3. c) First start with doing the problem in terms of the radius. Since the height of the cone is the same as the diameter, the volume of the cone is The volume of the hemisphere is The volume in terms of r for the figure is The height of the full figure T = 3r. Therefore Substituting in for r we get