course Mth 158 9-22-09 4:58 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** 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. ** ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same formula above, you would find the ^ of all three and add 2 of them to see if the sum equals one of the other legs. 10^2 = 100 24^2 = 576 26^2 = 676 Since the sum of the sqrt of 10 and the sum of the sqrt of 24 = 676, you conclude that the leg 26 is the hypotenuse of the right triangle. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think the given solution explained it a little better than I did, but we both had the same answer. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula is Volume = 4/3 * pi * r^3 Surface = 4 * pi * r^2 So if the radius is 3, you would use the above formula to obtain: v = 4/3 * pi * 3^3 = 4/3 * pi * 27 = 36 pi s = 4 * pi * 3^2 = 4 * pi * 9 = 36 pi confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 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 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I always forget to do the last step and raise it to the power of 3 for volume and 2 for surface area. I think that is the only discrepancy in the two solutions. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I still have trouble finding the solution to problems like this. I have read it in the book and just don’t grasp the meaning. This is one I will take to my tutor for further explanation. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 • area = pi r^2 = 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. ** "