course Mth 158 Am I right that there was no assignment 4? ʔӹ]ZyXfꏤdassignment #003
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13:35:52 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 --> c^2= a^2 + b^2 c^2= 14^2 = 48 ^2 C^2=2500 c=50
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13:36:19 ** 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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13:37:38 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 --> Yes, because 26^2=10^2 =24^2 When you work this out you get 676=676
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13:37:46 ** 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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13:38:54 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 --> A= 4 pi r^2 A= 4 (3.14) * 3^2 A= 113.04
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13:40:50 ** 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 --> I didn't get the sam answer but I think I just multiplies out my 4 pi.
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14:31:59 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 --> Substract the area of the pool 4 pi 10^2 from the area of the pool and deck 4 pi 11.5^2 to get the area of the deck alon. This substract out to be 1661- 1256= 405
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14:37:21 ** 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 --> I see itis easier to obtain the radius of the pool plus the deck and then work the appropriate equation
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14:38:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> There was no surprises one just has to visual the problem
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14:38:24 005. `query 4
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RESPONSE -->
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14:43:39 Query 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 --> 8 ( 4X^3 -3x^2-1) - 6(4 x^3 =8x -2) 32x^3- 24x^2 -8 -24x^3- 48x +12 8x^3 -24x^2 -48 x +4
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14:48:27 ** 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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RESPONSE --> i am corrected in that you group like terms together first.
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14:59:10 Query R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?
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RESPONSE --> (-2x- 3) 3-x (-2x-3) (-x+3) 2x^2 = 3x -9
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15:02:15 ** 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 --> Use the distributive law instead of FOIL
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15:05:43 Query 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-1) (x +1)= x^2 - 1^2 using the difference of two squares
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15:06:06 ** 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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RESPONSE --> OK
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15:13:54 Query 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 --> (2x=3y) ^ 2= (2x+2y)(2x + 3y) =4x^2 =3xy +9y^2
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15:17:13 ** 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 --> I missed multiplied 3y * 3y
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15:21:43 Query R.4.105 \ 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 --> When we are multiplying polynimals the term of the polynomial is multiplies and the exponents will add according to the basic rules of exponents.
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15:22:52 ** 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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RESPONSE --> OK
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15:23:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No comments
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