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Modules
Module 1. COMMON MATHEMATICAL GADGETS OR APPARATUSES
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Lesson 2. Manipulatives
Module 2: THE MATHEMATICAL TOOLS
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Lesson 1. Percentage Calculator
Lesson2 Binary, Decimal, Hexadecimal Converter
Lesson 3. Prime Factorization tool
Lesson4 Scientific Calculator and Straight line Graph Calculator
Lesson5 Which Day of the week You Were Born On
Lesson 6 Polyhedron Models
Lesson 7.Greatest Common factor Tool and Least Common Multiple Tool
Lesson 8 Quadratic Equation Solver
Activities
Activity #1. Let's Count It On!
Activity #2.Name Me
Activity #3.Right Triangle Problem
Activity #4.The Problem Pool Balls Puzzle
Activity #5. Gardens Puzzle
Activity #6. Meeting the Challenge
Activity #7. Verbal Problems
Activity #8."A Rare Math problem solving Course"
Activity #9.Math Patterns on a Number Wheel: Threes & Sevens
Activity #10. Hanayama Devil Puzzle Solution
Activity #11. if you are genius solve this
Activity #12. The Six Math Problem
Activity #13. A Geometry problem
Activity #14. The 97 Math Problem
Activity #15. Logical thinking
Activity #16. Geometry Problem
Activity #17. Sudoku Game
Activity #18. Algebra Problem
Activity #19. Mathematics Crossword Puzzle
Activity #20. Probability with dice
Answer Key
Activity #1. Let's Count It On!
Activity #2.Name Me
Activity #3.Right Triangle Problem
Activity #4.The Problem Pool Balls Puzzle
Activity #5. Gardens Puzzle
Activity #6. Meeting the Challenge
Activity #7. Verbal Problems
Activity #12. The Six Math Problem
Activity #13. A Geometry Problem Solution
Activity #14. The 97 Math Problem Solution
Activity #15. Logical thinking
Activity #16. Geometry Problem
Activity #17. Sudoku Game
Activity #18. Algebra Problem
Activity #19. Mathematics Crossword Puzzle
Profile
Geometry Problems
Use cosine law in triangle ABE: 132 = 202 + 92 - 2(20)(9)cos(T)
Use cosine law in triangle ACB: x2 = 202 + 92 - 2(20)(9)cos(90o - T)
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Note that cos(90o - T) = sin(T) and rewrite the second equation as
Use cosine law in triangle ACB: x2 = 202 + 92 - 2(20)(9)sin(T)
Solve the first equation for cos(T).
cos(T) = 13/15
Use trigonometric identity to find sin(T) = 2 sqrt(14) / 15
Substitute sin(T) by 2 sqrt(14) / 15 in the third equation and solve for x
x = sqrt( 481 - 48 sqrt( 14 ) ) = 17.4 (approximated to 3 significant digits)
Solve x - y = 1 for x (x = 1 + y) and substitute in the equation of the circle to obtain:
(1 + y)2 + 2·(1 + y) + y2 + 4y = -1
Write the above quadratic equation in standard form and solve it to obtain
y = - 2 + sqrt(2) and y = - 2 - sqrt(2)
Use x = 1 + y to find x
Points of intersection: ( -1 + sqrt(2), - 2 + sqrt(2) ) and ( -1 - sqrt(2) , -2 - sqrt(2) )
We first graph the lines y = x and y = -2x + 3 in order to locate the points of intersection of the lines and the x axis and identify the triangle in question.
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The height is the y coordinate of the point of intersection of the lines y = x and y = -2x + 3 found by solving the system of equations.
solve : y = -2x + 3 , y = x , solution: (1 , 1) which also the point of intersection. The y coordinate = 1 and is also the height.
The length of the base is the x intercept of the line y = -2x + 3 which is x = 3/2.
Area of the shaded triangle = (1/2)(1)(3/2) = 3/4
The formula for the area using two sides and the internal angle they make, may be written as follows
18 = (1/2) * 5 * 10*sin(A)
which gives: sin(A) = 18/25
We now use the cosine formula to fin the length x of the third side opposing angle A as follows:
x2 = 52 + 102 - 2*5*10*cos(A)
with cos(A) = sqrt(1 - sin(A)2)
Substitute in the expression for x2 and solve for x to obtain x = 7.46 (approximated to 3 significant digits)
http://www.analyzemath.com/high_school_math/grade_11/geometry.html