FORM 5 MATHEMATICS Chapter 3 TRANSFOMATION 111 NAME :__________________________________________ FORM 5 _________________________ 3.1 TRANSLATION (a) Base on the graph below‚ state the translation (i) A → A’ translation (ii) B → B’ (iii) C → C’ (iv) D → D’ (v) E → E’ (b) A Point P is located at coordinates (4‚3) on a Cartesian plane. P’ is the image of P under a translation
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area of cone = rl b hyp 1 3 h r opp adj = hyp cos opp = hyp sin opp = adj tan sin adj hyp tan or C opp hyp cos adj In any triangle ABC opp adj b a A B c a sin A Sine rule: b sin B c sin C Cosine rule: a2 = b2 + c2 – 2bc cos A Area of triangle = 1 2 ab sin C cross section lengt h Volume of prism = area of cross section length Area of a trapezium = r 1 2 (a + b)h a
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02.01 Classifying Triangles Warning: There is a checkbox at the bottom of the exam form that you MUST check prior to submitting this exam. Failure to do so may cause your work to be lost. ------------------------------------------------- Top of Form Question 1 (Multiple Choice Worth 2 points) What is the measure of the third angle? 30.5 55 35 149.5 Question 2 (Multiple Choice Worth 2 points) What are two qualities that make an equilateral triangle unique? Three congruent
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multiply Your students’ interest will with Microsoft® Mathematics 4.0. While teachers are under pressure to raise math and science test scores on standardized tests‚ many students find math their most frustrating subject. With Microsoft Mathematics‚ teachers can equip students with the tools they need to grasp the concepts behind the correct answers. When that happens‚ students’ engagement and comprehension can rise exponentially. MICROSOFT MATHEMATICS 1 Encourage. Visualize
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triggered off further efforts; further research was done to solve future surveying problems. It was not until on June 2013 while studying on curvilinear coordinates that we were prompted with an idea which enabled us to discover that the area of a triangle could be found by a relation connecting area to length of one side and trigonometric ratios of two angles emanating from the line considered. The knowledge of Vector analysis‚ equation of straight lines and calculus all combined proved to be
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Introduction to Epidemiology Module 1 SLP Trident University Dr. Narrad Beharry Due: February 24‚ 2014 According to the epidemiologic triad‚ there are three factors that influence the occurrence of disease: 1. Etiologic agent 2. Host factors 3. Environmental factors Describe each of these factors for a Salmonella foodborne outbreak. What are some of the host factors that can influence the occurrence of a disease? Salmonella Most people associate salmonella infection (salmonellosis)
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2 h 3 In anyCurved surface area of cone πrl triangle ABC – Area of triangle = 1 ab sin C 2 In any triangle ABC h b a Volume of prism = area of cross section × length a = b = c Sine rule : sin A sin B sinCross C section b of c Volume of a Sine Rule: prism = area cross section × length sin A In any triangle ABC sin B sin C Cosine rule: a2 = b2 + c2 – 2bc cos A Cross section c A c B b a l r r len Cosine Rule: a2 Area of triangle = 1
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You Can prove that radius to the point of contact of a tangent is perpendicular by Take two Iron Rings with a radius of a pin that you have like a stitching needle.. The iron rings should be at a thickness of 1 cm... Join the two rings by superimposing but not exactly‚ without gap. just leave a gap between the two circumference of the circle such that there is parallel gap throughout the two rings then join them at two points{one at any where and another at straight opp to the other} using m~seal
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w w om .c Paper 2 s er * 3 3 3 9 6 5 2 2 8 0 * MATHEMATICS (SYLLABUS D) ap eP m e tr .X w UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Ordinary Level 4024/22 May/June 2012 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Geometrical instruments Electronic calculator READ THESE INSTRUCTIONS FIRST Write your Centre number‚ candidate number and name on all the work you hand in
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the width. The length is 3 more than twice the width‚ so The area is 560‚ so Equation: Plug in and solve for W: Solution: Use the Quadratic Formula: Since the width can’t be negative‚ I get . The length is 2. The hypotenuse of a right triangle is 4 times the smallest side. The third side is . Find the hypotenuse and the smallest side. Representation: Let s be the smallest side and let h be the hypotenuse. By Pythagoras‚ The hypotenuse is 4 times the smallest side‚ so Equation: Plug into and
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