course MTH 158 I got very confused with this assignment - not the work - but the query. I'm not sure why questions from R.4 were in with R.3 - I did some of them, but I really hadn't studied that yet, so I was unsure exactly what I was doing. v֣c˸IJDֶassignment #003
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22:41:35 R.3.12 (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 of a right triangle you would use the Pythagorean Theorem, which is c^2 = a^2 + b^2 To determine the hypotenuse of this particular triangle you would subsitute 14 for a and 48 for b. c^2 = 14^2 + 48^2 And solve for c c^2 = 196 + 2304 c^2 = 2500 `sqrt(c^2) = `sqrt(2500) c = 50 The hypotenuse of a right triangle with legs 14 and 48 is 50. confidence assessment: 3
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22:41:42 01-19-2008 22:41:42 ** 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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NOTES -------> ok
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22:44:37 R.3.18 (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 --> In order for a triangle to be considered a right triangle the square of the two legs must equal the square of the hypotenuse. In this case, if 10^2 + 24^2 = 26^2 then it is a right triangle. c^2 = a^2 + b^2 c^2 = 10^2 + 24^2 c^2 = 100 + 576 c^2 = 676 `sqrt(c^2) = `sqrt(676) c = 26 This is a right triangle and the hypotenuse is 26. confidence assessment: 3
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22:44:50 01-19-2008 22:44:50 ** 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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NOTES -------> ok
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22:48:49 R.3.30 (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 a sphere of radius 3 feet you would use the following formula and solve for V. V = (4/3) `pi (r^3) V = (4/3) `pi (3^3) V = (4/3) `pi (27) V = 36 `pi meters^3 V = 113.097 meters^3 To find the surface area of a sphere with radius 3 meters you would use the following formula and solve for S. S = 4 `pi r^2 S = 4 `pi 3^2 S = 4 `pi 9 S = 36 `pi meters^2 S = 113.097 meters^2 confidence assessment: 3
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22:49:29 01-19-2008 22:49:29 ** 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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NOTES -------> Was it just a coincidence that the both of these equatiosn equaled 36 or is that common when using these two formulas?
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22:54:47 R.3.42 (was R.3.36). A pool of radius 10 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 --> Because the diameter of the pool is 20 feet, the radius is 10 feet. The deck would add an additional 6 feet to the diameter of the pool, making the radius of the pool and deck 13 feet. The area of the deck and pool is A = `pi (r^2) A = `pi (13 ^ 2) A = 169 `pi feet^2 The area of the pool alone is A = `pi (r^2) A = `pi (10^2) A = 100 `pi feet^2 To find the area of deck, you must subtract the area of the pool from the area of the deck and pool together. Area of both - Area of pool = Area of deck 169 `pi - 100 `pi = 69 `pi feet^2 confidence assessment: 3
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22:56:09 01-19-2008 22:56:09 ** 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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NOTES -------> Ok.
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22:56:26 005. `query 5
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RESPONSE --> Not sure what this is self critique assessment: 0
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22:58:08 R.4.36 (was R.5.30). What is the single polynomial that is equal to 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 )?
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RESPONSE --> This kind of problem was not covered in this section. confidence assessment: 0
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22:58:26 01-19-2008 22:58:26 ** 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 can't isolate them like that. 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 **
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NOTES -------> o
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23:00:03 R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?
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RESPONSE --> 2x^2 - 3x - 9 But this was not covered in R.3 confidence assessment: 3
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23:00:19 ** 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. **
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RESPONSE --> ok self critique assessment:
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23:01:06 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?
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RESPONSE --> x^2 + 1 confidence assessment: 2
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23:01:42 01-19-2008 23:01:42 ** 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. **
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NOTES -------> I meant to say -1, this was not covered in R.3
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23:03:01 R.4.84 (was R.5.78). What is (2x + 3y)^2 and how did you obtain your result using a special product formula?
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RESPONSE --> 4x^2 + 4x6y + 9 Not covered in R.3 confidence assessment: 2
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23:03:23 ** The Special Product is (a + b)^2 = a^2 + 2 a b + b^2. Letting a = 2x and b = 3y we get (2x)^2 + 2 * (2x) * (3y) + (3y)^2, which we expand to get 4 x^2 + 12 x y + 9 y^2. **
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RESPONSE --> k self critique assessment:
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23:03:58 R.4.90 (was R.5.102). Explain why the degree of the product of two polynomials equals the sum of their degrees.
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RESPONSE --> I'm really not sure why R.4 is in with the R.3 problems? Did I mess something up? confidence assessment: 0
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23:04:13 01-19-2008 23:04:13 ** 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. **
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NOTES -------> ok
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23:04:20 01-19-2008 23:04:20 Add comments on any surprises or insights you experienced as a result of this assignment.
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NOTES -------> ok
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