course Mth158 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: Plug the given lengths in for a^2+b^2=c^2. 10^2+24^2=26^2 100 + 576= 676 676 = 676 The given lengths form a right triangle. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Using the Pythagorean Theorem we have c^2 = a^2 + b^2, if and only if the triangle is a right triangle. Substituting we get 26^2 = 10^2 + 24^2, or 676 = 100 + 576 so that 676 = 676 This confirms that the Pythagorean Theorem applies and we have a right triangle. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: Use the equation for volume and substitute the radius of 3 meters in for R. Volume = 4/3 * pi * R^3 V = 4/3 * pi * (3 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Use the equation for surface area below and substitute the radius 3 meters in for R. Surface Area = 4 * pi * R^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** To find the volume and surface are a sphere we use the given formulas: Volume = 4/3 * pi * r^3 V = 4/3 * pi * (3 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):0 ------------------------------------------------ Self-critique rating #$&*0 ********************************************* 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: The pool would have a radius of 10 ft and 13 ft with the deck added in. The equation A=pi*r^2 can be used here where A=pi*13^2 A=pi*169ft ^2 I am unaware how to solve for the area of the deck. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Think of a circle of radius 10 ft and a circle of radius 13 ft, both with the same center. If you 'cut out' the 10 ft circle you are left with a 'ring' which is 3 ft wide. It is this 'ring' that's covered by the deck. The 10 ft. circle in the middle is the pool. 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. ** Self critique: I understand that by squaring the radius of the pool alone I am left with 100 pi ft^2 and can subtract this from the total of 169pi ft^2 to get an area for my deck of 69 pi ft^2 "