course Mth 152 ?W??????????d?assignment #024024. `query 24
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13:39:36 **** query 9.5.12 vol of sphere diam 14.8 **** What is the volume of the sphere and how did you obtain it?
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RESPONSE --> (4/3)(3.14)(7.4)^3 = 1696.54 cm^3
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13:39:42 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
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RESPONSE --> ok
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13:40:28 **** query 9.5.18 pyramid 12 x 4 altitude 10 **** What is the volume of the pyramid and how did you find it?
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RESPONSE --> (1/3)*12*4*10 = 160 ft.^3
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13:40:32 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
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RESPONSE --> ok
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13:41:52 **** query 9.5.24 bottle 3 cm alt 4.3 cm **** What is the volume of the bottle and how did you find it?
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RESPONSE --> (3.14)(1.5)^2(4.3) = 30.38 cm^3
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13:41:56 ** 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 **
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RESPONSE --> ok
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13:44:04 **** query 9.5.36 sphere area 144 `pi^2 **** What are the radius, diameter and volume of the sphere and how did you find them?
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RESPONSE --> r = sqrt.(144pi/pi/4) = 6m d = 2*6 = 12m V = (4/3)pi*6^3 = 288pi m^3
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13:44:21 ** 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)^2 = 48 pi m^2. **
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RESPONSE --> ok
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13:45:30 **** 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?
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RESPONSE --> 245pi = (1/3)pi*15*r^2 x = 7
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13:45:36 ** 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 / (3.14 * 15) ) = 3.9 approx. **
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RESPONSE --> ok
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13:46:05 **** query 9.5.51 plane intersects sphere passing 7 in from center forming circle with area 576 `pi **** What is the volume of the sphere and how did you obtain it?
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RESPONSE --> (62500/3)pi in.^3
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13:46:14 ** 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 the 24 in to the rim of the circle, 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. **
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RESPONSE --> ok
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13:46:22 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Alright
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13:46:29 I don't think that my last answer is correct. Also, in the first problem, the answer should be 1696.54 cm^3, instead of just 1696.54
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RESPONSE --> ok
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