course MTH 158 ËÕçϨâwƒh›ê´±¹á»¹¬°dú‹õÁéFªæªassignment #003
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20:33:19 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= 196+2304 C^2= 2500 C=sqrt 2500 C= 50
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20:33:52 ** 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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20:38:37 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 it is a right triangle. 26 would be the longest side which is the hypotenuse. so we square the sides to see if they equal 26^2 10^2=100 24^2=576 26^2=676 100+576=676
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20:38:48 ** 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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20:53:56 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 --> Volume of a sphere 4/3pi r^3 Surface area of a sphere 4pi r^2 V=4/3 pi 3m^3 V=4/3*3m^3*pi V=12m^3/3 pi V=4 pi meters^3 S=4 pi 3m^2 S= 12 pi meters^2
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20:56:46 ** 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 see the mistakes I made in this problem. I didnt do my exponents first. I totally confused myself.
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21:09:50 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 --> Area of deck = 69 pi ft^2 First I found the diameter of the entire pool and deck. 20+3+3=26' diameter I then found the area of the whole circle A=pi*13^2 A=169 pi ft^2 Then I found the area of the swimming pool A=pi * 10^2 A= 100 pi ft^2 Then I subtracted the area of the swimming pool from the entire circle to give me the area of the deck 169 pi ft^2-100 pi ft^2= 100 pi ft^2
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21:10:54 ** 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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21:12:10 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Just that I need to be careful with order of operations, take my time and do them in order.
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21:12:26 005. `query 4
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¥û}®öË™„¿ÆÊáý¢¬÷bß½Ÿ{Þ¯Æ˜Æ assignment #003 003. `query 3 College Algebra 06-01-2009
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21:15:21 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 -->
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21:15:24 ** 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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21:15:27 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 -->
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21:15:30 ** 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 -->
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21:15:34 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 -->
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21:15:38 ** 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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21:15:41 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 -->
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21:15:44 ** 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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21:15:46 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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