In the AUL CSI teams case #46787-35 where Charles Drummer A.K.A snake a local gang ring leader’s body exactly 35.38 ft. from an open window on the 35th floor of a high rise building which we measured was 320 ft. from the ground. The only lead we had was a supposed witness phone call somewhere in the area about a person jumping from a building. We knew the location of the cell phone tower that received the call, the time of the call, that the distance between the cell phone tower to the satellite
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trapezium = 2 (a + b)h a cross section Volume of prism = area of cross section × length h lengt h b Volume of sphere = 4 r3 3 Surface area of sphere = 4 r2 r Volume of cone = 1 r2h 3 Curved surface area of cone = rl l r h In any triangle ABC b A c C a B The Quadratic Equation The solutions of ax2 + bx + c = 0 where a ≠ 0, are given by
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area. Investigate the shape/s of the plot of land that have the maximum area. Solution -------- Firstly I will look at 3 common shapes. These will be: ------------------------------------------------------ [IMAGE] A regular triangle for this task will have the following area: 1/2 b x h 1000m / 3 - 333.33 333.33 / 2 = 166.66 333.33² - 166.66² = 83331.11 Square root of 83331.11 = 288.67 288.67 x 166.66 = 48112.52² [IMAGE]A regular square for this
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| | |MIDTERM - 2nd SEMESTER [pic] | |Subject: |Mathematics for Business |Professor: |Ricardo Torres | |Timing: |2 hours |Student’s name: |
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right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector likeA in Figure, we may wish to find which two perpendicular vectors, Ax and Ay, add to produce it. The vector A, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, Ax and Ay. These vectors form a right triangle. The analytical
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not indicate in which direction the 2x + 9 paces should go, it can be assumed that his and Buckwheat’s paces should end up in the same place. When sketched on scratch paper, a right triangle is formed with 2x + 9 being the length of the hypotenuse, and x and 2x + 6 being the legs of the triangle. When a right triangle is involved, the Pythagorean Theorem helps solve for x.
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BELL RINGERS – THE NUMBER SYSTEM – GRADE 8 Task 1 - Which of the following numbers are perfect squares? Explain your reasoning. 5, 9, 36 and 41 Task 2 - Which of the following numbers are perfect cubes? Explain your reasoning. 8, 12, 27 and 64 Task 3 - Problem: Represent the following rational number in fraction form 0.333? Task 4 - Instructions: Decide whether each of the following numbers is rational or irrational. 1. 0.333 ______________________ 2
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Learning Activity 5: Paper Works (Hand-outs Use) Name of FS Student: Geraldine S. Sitoy Course: BTTE Year & Section: 4A Resource Teacher: Cooperating School: Grade Level Observed: My Goal At the end of this activity, I will gain competence in making instructional materials (hand-outs) appropriate to the learning content. My Tasks I am going to choose one or two hand-outs used by the teacher in her lesson. Analyze the hand-outs in terms of its contents, learning activities, and assessment plan
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program will calculate the area of a right triangle. The program will ask the user to enter the base and height and then use these values to calculate and then print the area of the triangle. If the area of the triangle is greater than 100 square units, an additional message is printed stating the triangle is too large for the specification. However; if the triangle is less than or equal to 100 square units, the additional message will state the triangle is within specifications. The design step will
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prism = area of cross-section × length h lengt r Volume of sphere = 4 πr3 3 Surface area of sphere = 4πr2 l Volume of cone = 1 πr2h 3 h Curved surface area of cone = πrl r In any triangle ABC Sine rule C a = b = c sin B sin C sin A Area of triangle = 1 ab sin C 2 a b Cosine rule a2 = b2 + c2 – 2bc cos A A c The Quadratic Equation The solutions of ax2 + bx + c = 0 where a ≠ 0 are given by (185-13) x= −b ± (b 2 − 4 ac ) 2a
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