course mth 158 Question: will we need to have memorized all of the geometrical formulas for the test?
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21:26:50 query R.3.16 \ 12 (was R.3.6) What is the hypotenuse of a right triangle with legs 14 and 48 and how did you get your result?
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RESPONSE --> To find the hypotenuse (c): a^2 + b^2 = c^2 14^2 + 48^2 = c^2 196 + 2304 = c^2 2500 = c^2 sqrt2500 = c 50 = c
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21:27:02 ** The Pythagorean Theorem tells us that c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse. Substituting 14 and 48 for a and b we get c^2 = 14^2 + 48^2, so that c^2 = 196 + 2304 or c^2 = 2500. This tells us that c = + sqrt(2500) or -sqrt(2500). Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. **
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RESPONSE --> ok
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21:29:55 query R.3.22 \ 18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer?
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RESPONSE --> You would use the Pythagorean Theorem to find out if this is a right triangle. c^2 = a^2 + b^2 26^2 ? 10^2 + 24^2 676 ? 100 + 576 676 = 676 This triangle is a right triangle
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21:30:03 ** Using the Pythagorean Theorem we have c^2 = a^2 + b^2, if and only if the triangle is a right triangle. Substituting we get 26^2 = 10^2 + 24^2, or 676 = 100 + 576 so that 676 = 676 This confirms that the Pythagorean Theorem applies and we have a right triangle. **
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RESPONSE --> ok
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21:35:29 query R.3.34 \ 30 (was R.3.24). What are the volume and surface area of a sphere with radius 3 meters, and how did you obtain your result?
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RESPONSE --> To find the volume of the sphere: V = 4/3 pi r^3 = 4/3 (3.14) (3^3) = 4/3 (84.78) = 113.09 ft^3 To find the surface area of a sphere: SA = 4 pi r^2 = 4 (3.14)(3^2) = 4 (28.27) = 113.09 ft^2
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21:36:30 ** To find the volume and surface are a sphere we use the given formulas: Volume = 4/3 * pi * r^3 V = 4/3 * pi * 3^3 V = 4/3 * pi * 27 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * 3^2 S = 4 * pi * 9 S = 36pi m^2. **
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RESPONSE --> not sure if I understand V = 36pi m^3
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21:42:12 query R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result?
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RESPONSE --> 20 ft = diameter of pool 3 ft = diameter of deck First, find the area of the deck: A = pi r^2 A = pi (20^2) A = 1256.64 ft^2 20 + 3 = diameter of pool and deck A = pi r^2 = pi (23^2) = 1661.90 ft^2 = area of pool & deck Then subtract area of pool from pool+deck to get area of deck: 1661.90 ft^2 - 1256.64 ft^2 = 405.26 ft^2
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21:44:03 ** The deck plus the pool gives you a circle of radius 10 ft + 3 ft = 13 ft. The area of the deck plus the pool is therefore pi * (13 ft)^2 = 169 pi ft^2. So the area of the deck must be deck area = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2. **
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RESPONSE --> wrong -- computed diameter instead of radius Forgot to divide by 2
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21:45:37 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found it hard to think of logical equations in terms of geometry. For example, the sight distance from a tower. I would have never thought to do the geometry based on the radius of the earth to find that. Once you understand the concept, it seems easy.
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