course Mth 151 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: A^2 + B^2 = C^2 10^2 + 24^2 = 26^2 100 + 576 = 676 676= 676 Yes by using pyth. Theorom confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Using the Pythagorean Theorem we have c^2 = a^2 + b^2, if and only if the triangle is a right triangle. Substituting we get 26^2 = 10^2 + 24^2, or 676 = 100 + 576 so that 676 = 676 This confirms that the Pythagorean Theorem applies and we have a right triangle. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* 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: V = 4/3 * pi * r^3 V= 4/3 * pi *3^3 V= 4/3 * pi * 27 V=36 meters ^3 S= 4 * pi * r^2 S= 4 * pi * 3^2 S= 4 * pi * 9 S= 36 pi meters ^2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** To find the volume and surface are a sphere we use the given formulas: Volume = 4/3 * pi * r^3 V = 4/3 * pi * (3 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* 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: area = pi r^2 area = pi * 13^2 area= pi * 169 area= 169 pi ft ^2 area of deck = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Think of a circle of radius 10 ft and a circle of radius 13 ft, both with the same center. If you 'cut out' the 10 ft circle you are left with a 'ring' which is 3 ft wide. It is this 'ring' that's covered by the deck. The 10 ft. circle in the middle is the pool. The deck plus the pool gives you a circle of radius 10 ft + 3 ft = 13 ft. The area of the deck plus the pool is therefore • area = pi r^2 = pi * (13 ft)^2 = 169 pi ft^2. So the area of the deck must be • deck area = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2. ** "
course Mth 151 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** ERRONEOUS STUDENT SOLUTION: To make this problem into a single polynomial, you can group like terms together. (8-6)+ (4x^3-4x^3) + (-3x^2) + (8x) + (-1+2). Then solve from what you just grouped...2 (-3x^2+8x+1). INSTRUCTOR CORRECTION: 8 is multiplied by the first polynomial and 6 by the second. You need to follow the order of operations. Starting with 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 ) use the Distributive Law to get 32 x^3 - 24 x^2 - 8 - 24 x^3 - 48 x + 12. Then add like terms to get 8x^3 - 24x^2 - 48x + 4 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: * R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-2x - 3) (3 – x) = -2x (3 - x) - 3 (3 - x) = -2x (3) - 2x (-x) - 3 * 3 - 3 (-x) = -6x + 2 x^2 - 9 + 3x = 2 x^2 - 3 x - 9 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Many students like to use FOIL but it's much better to use the Distributive Law, which will later be applied to longer and more complicated expressions where FOIL does not help a bit. Starting with (-2x - 3) ( 3 - x) apply the Distributive Law to get -2x ( 3 - x) - 3 ( 3 - x). Then apply the Distributive Law again to get -2x(3) - 2x(-x) - 3 * 3 - 3 ( -x) and simiplify to get -6x + 2 x^2 - 9 + 3x. Add like terms to get 2 x^2 - 3 x - 9. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: * R.4.66 (was R.5.60). What is the product (x - 1) ( x + 1) and how did you obtain your result using a special product formula? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-1) (x+1) = x (x + 1) - 1 (x+1) = x*x + x * 1 - 1 * x - 1 * 1 = x^2 +- x - x + - 1 = x^2 - 1 I used the Distributive Law confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with (x-1)(x+1) use the Distributive Law once to get x ( x + 1) - 1 ( x+1) then use the Distributive Law again to get x*x + x * 1 - 1 * x - 1 * 1. Simplify to get x^2 +- x - x + - 1. Add like terms to get x^2 - 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: * R.4.84 (was R.5.78). What is (2x + 3y)^2 and how did you obtain your result using a special product formula? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** The Special Product is • (a + b)^2 = a^2 + 2 a b + b^2. Letting a = 2x and b = 3y we substitute into the right-hand side a^2 + 2 a b + b^2 to get (2x)^2 + 2 * (2x) * (3y) + (3y)^2, which we expand to get 4 x^2 + 12 x y + 9 y^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sure how to get this answer or which formula to use. ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: * R.4.105 \ 90 (was R.5.102). Explain why the degree of the product of two polynomials equals the sum of their degrees. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is true because of the Distributive Law which states that you multiply the highest power term in the first polynomial by the highest power term in the second. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** STUDENT ANSWER AND INSTRUCTOR COMMENTS: The degree of the product of two polynomials equals the sum of their degrees because you use the law of exponenents and the ditributive property. INSTRUCOTR COMMENTS: Not bad. A more detailed explanation: The Distributive Law ensures that you will be multiplying the highest-power term in the first polynomial by the highest-power term in the second. Since the degree of each polynomial is the highest power present, and since the product of two powers gives you an exponent equal to the sum of those powers, the highest power in the product will be the sum of the degrees of the two polynomials. Since the highest power present in the product is the degree of the product, the degree of the product is the sum of the degrees of the polynomials. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):