#$&* course Mth 152 Nov 29 - 10:19pm 024. ``q Query 24
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Given Solution: `a I use the formula for finding the volume of a sphere which is 4/3(3.14)(r^3). Since the diameter is 14.8, the radius is half that which is 7.4. V = 4/3 * 3.14 * 7.4^3 V = 4/3 * 3.14 * 405.224 V = 1696.54 The volume is 1696.54 cm^3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.5.18 pyramid 12 x 4 altitude 10 **** What is the volume of the pyramid and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A pyramid 12 feet by 4 feet with a 10-foot height or altitude/ To find the volume of a pyramid you multiply base * height * 1/3. To get the base we multiply 12*4 = 48 Then we put the right numbers into the formula. 48*10*1/3 = 480*1/3 = 160ft^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a I used the formula : V = 1/3Bh The base = 12 * 4 = 48 V = 1/3 * 48 * 10 V = 1/3 * 480 V = 160 The volume is 160ft.^3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.5.24 bottle 3 cm alt 4.3 cm **** What is the volume of the bottle of typewriter correction fluid, diameter 3 cm and height 4.3 cm and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a circular cylinder like problem #13 you use the formula: h*3.14*r^2 3.14*1.5^2*4.3cm 3.14*2.25*4*3 30.38cm^3 (because the dimension is 3) confidence rating #$&*: okay ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** The figure is a right circular cylinder with V = 3.14 * r^2 * h Since the diameter is 3, then the radius is 1.5 V = 3.14 * 1.5^2 * 4.3 V = 3.14 * 2.25 * 4.3 V = 30.38 cm^3 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay Self-critique Rating: Okay ********************************************* Question: `q Query 9.5.36 sphere area 144 `pi m^2 **** What are the radius, diameter and volume of the sphere and how did you find them? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We need to find the volume of a sphere and from the information table all we know is that it’s surface area is 144’pi m^2. We need to find the radius, the diameter, and the volume. The surface area of a sphere is found by 4 pi r 2 which in our problem = 144’pi m^2. If we divide 144 by 4 we get 36. So r^2= 36m^2. 36m squared would give us 6m which would be the radius. If we know the radius we know the diameter because it’s just the radius doubled or 6*2= 12m. The formula for volume is 4/3 * 3.14 *6 = 288’pi m^3 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** Sphere area is 4 pi r^2, so we have 4 pi r^2 = 144 pi m^2. Dividing by 4 pi we get r^2 = 36 m^2. Taking the square root of both sides we get r = 6 m. From this we find that the diameter is 2 * 6 m = 12 m and the volume is 4/3 pi * (6 m)^3 = 288 pi m^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.5.48 cone alt 15 rad x vol 245 `pi **** what is the value of x and how did you find your result? V=254pi A right circular cone 15 radius YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a right circular cone you use the formula 1/3* 3.14 * r^2 * h 1/3 *3.14 *15^2 I’m not sure where to go from here confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** We have V = 1/3 pi r^2 h. To solve for r we multiply both sides by 3 / (pi * h) to get 3 V / (pi * h) = r^2 then take the square root to get r = sqrt(3 V / ( pi * h) ). Substituting we get r = sqrt( 3 * 245 pi / (pi * 15) ) = sqrt(49) = 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why do we multiply both sides by 3? ------------------------------------------------ Self-critique Rating: 1
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Given Solution: `a ** The circle does have radius sqrt(576 in^2) = 24 in. However that is not the radius of the sphere since the plane containing the circle passes 7 in from the center of the sphere. So the center of the circle is not the center of the sphere. The center of the circle is 7 in from the center of the sphere. Note also that a line from the center of the sphere to the center of the circle will be perpendicular to the plane of the circle. Thus if you start at the center of the sphere and move the 7 in straight to the center of the circle, then move along the plane of intersection (in any direction) for 24 in (at which point you encounter the rim of the circle, whose radius you recall is 24 in), then back to the center of the sphere you will have traced out a right triangle with legs 7 in and 24 in. The hypotenuse of the triangle is the radius R of the sphere. So we have R^2 = 7^2 + 24^2 = 625 and R = 25. The radius of the sphere is 25 in. Its volume will therefore be 4/3 pi r^3: V = 4/3 pi r^3 = 4/3 pi * (25 in)^3 = 4/3 pi * 16000 in^3 (approx.) = 67000 in^3, approx. ** STUDENT COMMENT: I don't understand how you drew a triangle inside of the circle. How did you know to do that instead of adding the 7 to the radius of the plane? INSTRUCTOR RESPONSE: The triangle wasn't inside the circle. It was, however, inside the sphere. Its 7-inch leg runs from the center of the sphere to the center of the circle. Its 24 in leg runs from the center of the circle to any point on the circle. That point is also on the sphere. A radius of the sphere runs from the center of the sphere to any point on the sphere. The hypotenuse of the right triangle runs from the center of the circle to a point on the sphere, so the hypotenuse is a radius of the sphere. The key part of the explanation is quoted in the paragraph below. You can probably get by OK without completely understanding this (nothing this challenging on the test), but if you still have trouble with the given solution and want to get to the bottom of it, then tell me, phrase by phrase, what you do and do not understand in the above explanation and in this paragraph: 'Thus if you start at the center of the sphere and move the 7 in straight to the center of the circle, then move along the plane of intersection (in any direction) for 24 in (at which point you encounter the rim of the circle, whose radius you recall is 24 in), then back to the center of the sphere you will have traced out a right triangle with legs 7 in and 24 in. The hypotenuse of the triangle is the radius R of the sphere.' Self-critique : How did you get 16000? ------------------------------------------------ Self-critique Rating: okay