#$&* course Mth 158 2/1 2:30pm 003. `* 3
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Given Solution: * * ** The Pythagorean Theorem tells us that c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse. Substituting 14 and 48 for a and b we get c^2 = 14^2 + 48^2, so that c^2 = 196 + 2304 or c^2 = 2500. This tells us that c = + sqrt(2500) or -sqrt(2500). • Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. ** ********************************************* Question: * R.3.22 \ 18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it is a right triangle The Pythagorean Theorum states that the sides of a right triangle equal a^2 + b^2 = c^2 10^2 + 24^2= 26^2 100+576 = 676 Therefore, a triangle with sides 10, 24, and 26 is a right triangle. confidence rating #$&*: 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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: The Formula for the volume of a sphere V=4/3 pi r^3 Substitute the value of the radius for r V= 4/3 pi (3m)^3 V=4/3 pi 27m^3 V= 36 pi m^3 The formula for the surface area of a sphere S= 4 pi r^2 Substitute the value of the radius for r S= 4 pi (3m)^2 S= 4 pi 9m^2 S= 36 pi m^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** To find the volume and surface are a sphere we use the given formulas: Volume = 4/3 * pi * r^3 V = 4/3 * pi * (3 m)^3 V = 4/3 * pi * 27 m^3 V = 36pi m^3 Surface Area = 4 * pi * r^2 S = 4 * pi * (3 m)^2 S = 4 * pi * 9 m^2 S = 36pi m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First the area of the pool should be determined A=pi r^2 A= 100 pi ft^2 Now the area of the deck plus the area of the pool We know the radius of the deck is 10 ft + 3ft so A= pi (13ft)2 A= 169 pi ft^2 Now in order to determine the area of the deck itself, we have to subtract from it the area of the pool. 169 pi ft^2 - 100 pi ft^2= 69 pi ft^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ The only reason I knew how to do this problem was because of the homework assignments on mathxl. I love the program since it allows me to work through my mistakes, and tells me why my answers are wrong and which direction I should go to fix them. They also provide information that is great to add to my notes.
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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 (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: First the area of the pool should be determined A=pi r^2 A= 100 pi ft^2 Now the area of the deck plus the area of the pool We know the radius of the deck is 10 ft + 3ft so A= pi (13ft)2 A= 169 pi ft^2 Now in order to determine the area of the deck itself, we have to subtract from it the area of the pool. 169 pi ft^2 - 100 pi ft^2= 69 pi ft^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ The only reason I knew how to do this problem was because of the homework assignments on mathxl. I love the program since it allows me to work through my mistakes, and tells me why my answers are wrong and which direction I should go to fix them. They also provide information that is great to add to my notes.
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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 (if necessary): ------------------------------------------------ Self-critique rating: #*&!