course Mth 152 I have been helping someone move all week outside of work so its been difficult for me to take the time to get to the computer (my paper work part of homework I do up at work)... please ignore the dates of these as the time clock on the pc I have been working on to do these is failing. У
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00:28:26 **** query 9.3.12 area of parallelogram **** What is the area of the given parallelogram and how did you obtain it?
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RESPONSE --> I got the area of the parallelogram by multiplying the height times the base. 4*2.5=10 (inches)
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00:28:29 The formula for the area of a parallelogram is A = bh ( area = base times height) From the figure the base is 4 and the altitude is 2.5 so we have A = 4 * 2.5 = 10. The area is 10 in.
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RESPONSE --> Yes
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00:32:31 **** query 9.3.18 area of trapezoid **** What is the area of the given trapezoid and how did you obtain it?
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RESPONSE --> I was able to get the area of the trapezoid by multiplying 1/2 the height by the sum of the two bases... Area= .5(3) (4+5) or 1.5(9)= Area= 13.5
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00:32:36 The formula for finding the area of a trapezoid is A = 1/2h ( b + B ) h = 3, b = 4, B = 5 A = 1/2 (3) (4 + 5) A = 1/2 (3) (9) A = 1/2 (27) A = 13.5 The area is 13.5
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RESPONSE --> yes
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00:36:28 **** query 9.3.24 dim of rect with lgth 20 more than wdth and perimeter 176 **** What are the dimensions of the rectangle?
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RESPONSE --> I solved this by using this formula to find the dimensions.... P= 2x (width)+ 2y (length) 176= 2(x)+ 2(20+x) 176= 4x+20 176-20=4x+20-20 166=4x 166/4= 41.5 the width=41.5 and the length is 61.5
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00:36:47 ** The perimeter of a rectangle is P = 2l + 2w, where l and w are length and width. From the given information Length = 20 + w Perimeter = 176 This gives us the equation 176 = 2 (20 + w) + 2w which we proceed to solve for w: 176 = 40 + 2w + 2w by the Distributive Law. Simplifying we get 176 = 40 + 4w Subtract 40 from both sides to get 136 = 4w Divide both sides by 4 to get w = 34 l = w + 20 so l = 34 + 20 = 54 We have Length = 54 and Width = 34 Checking, we have perimeter = 2 * length + 2 * width so we should have 176 = 2(54) + 2(34). The right-hand side does give us 176 so the solution checks out. **
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RESPONSE --> Okay, I see what I did I think...
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00:39:11 **** query 9.3.48 trap bases x, x+4 alt 3 area 30 **** What is the value of x?
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RESPONSE --> I used the following formula... Area= 1/2height (base1+base2) 30= 1.5 (b+ (b+4) 30= 1.5x+ 1.5x+ 6 30= 3x+6 24= 3x x= 8
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00:39:15 The formula for finding the area of a trapezoid is A = 1/2h ( b + B). We have A = 30 h = 3 B = x + 4 b = x 30 = 1/2(3) ( x + x+4) 30 = 1.5 ( x+x+4) 30 = 1.5x + 1.5x + 6 30 = 3x + 6 Subtract 6 from both sides 24 = 3x Divide both sides by 3 x = 8 The answer is 8.
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RESPONSE --> yes
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00:41:27 **** query 9.3.54 $60 to paint ceiling of 9 x 15 rm, how much to paint if dimensions 18 x 30 **** What is the cost for the second ceiling? **** How did you use the results of exercise 53 to obtain your result?
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RESPONSE --> This was easy as the dimensions, width and length, were doubled... as such, this increases the total square space coverage by 4 (I know that from designing Dungeons & Dragons products and dealing with the battle grids and dungeon/castle maps a great deal.... plus village layouts)... Simply take the original cost then and multiply it by 4 for the total area= $240.
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00:41:30 The sides doubled from 9 ft. to 18 ft. and from 15 ft. to 30 ft. When the sides are doubled the area increases by a factor of 4. So the cost is $60 * 4 = $240 The cost for the second ceiling is $240
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RESPONSE --> Yes
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00:43:45 **** query 9.3.60 triangle alt 9 on top of 10 x 4 rect; parallelogram alt 3 under **** What is the area of the given figure and how did you obtain your result?
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RESPONSE --> I used the formula for a triangle area--- 1/2bh A= 1/2 (10) (9) A= 1/2 (10*9) A= 1/2 (90) A= 45 Then I used the formula for recs... A= (10)(4) Area= 40 Then I used the form. for para... A= (10) (3) Area= 30 Working them all together we get: 45+40+30=115
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00:43:48 ** Here's a solution by a student from a previous semester: I found the area of each figure and then added those three results together. Area of Triangle = 1/2bh A = 1/2 (10) (9) A = 1/2 (90) A = 45 Area of Rectangle = lw A = (10) (4) A = 40 Area of Parallelogram = bh A = (10) (3) A = 30 45 + 40 + 30 = 115 Area = 115 **
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RESPONSE --> Yes
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00:51:46 **** query 9.3.67 26 m diam circle inscribed in sq; area outside circle **** What is the area of the shaded region and how did you obtain it?
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RESPONSE --> As the circle has a diameter of 26m its radius is 1/2, or 13m... thus the area would be.... A= pi r^2= pi (13)^2= 169pi^2= 531^2 There area of the square... A= (26)^2=676^2 Area of the shaded area. 676^2-531^2= 145^2
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00:51:50 ** The circle has diameter 26 m so its radius is 13 m and its area is A = pi r^2 = pi * (13 m)^2 = 169 pi m^2 = 531 m^2. The area of the square is the square of its side A = (26 m)^2 = 676 m^2. The area of the shaded region is the difference } 676 m^2 - 531 m^2 = 145 m^2. **
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RESPONSE --> Yes, more or less
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01:12:09 **** query 9.3.72 10, 12, 14 in pizzas for 11.99, 13.99, 14.99 **** Which pizza is the best buy and how did you
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RESPONSE --> I went ahead and measured the pizzas as a circle.... (assuming these aren't square pan pizzas) Area = 3.14 (5) ^ 2 = 78.5 Area = 3.14 (6) ^ 2 = 113.04 Area = 3.14 (7) ^ 2 = 153.86 After that I divided the prices by the areas... $11.99 / 78.5 = 15.3 $13.99 / 113.04 = 12.4 $14.99 / 153.86 = 9.7 Given the results the 14 inch pizza is the best choice
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01:12:17 obtain your result?
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RESPONSE --> Eh?
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01:12:20 Student Solution: The thickness is about the same for all three pizzas so the amount of pizza can be measured by its area. I therefore found the area of each. A = 3.14 * r^2 and radius is 1/2 the circumference. So we get areas A = 3.14 (5)^2 = 78.5 A = 3.14 (6)^2 = 113.04 A = 3.14 (7)^2 = 153.86 I then divide the prices and these answers to get the price per square inch. $11.99 / 78.5 = 15.3 $13.99 / 113.04 = 12.4 $14.99 / 153.86 = 9.7 Since 9.7 is the least, and since this is the result for the 14 inch pizza, the 14 in. pizza for $14.99 is the best buy.
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RESPONSE --> Yes
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01:12:33 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Not a problem with this one
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01:12:35 There weren't any big surprises.
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RESPONSE -->
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" course Mth 152 I have been helping someone move all week outside of work so its been difficult for me to take the time to get to the computer (my paper work part of homework I do up at work)... please ignore the dates of these as the time clock on the pc I have been working on to do these is failing. ίřweՇ
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14:18:35 **** query 9.4.6 ABC, DEF transversed by EOB at rt angles; OB = EO; show triangles ABO and DOF congruent.
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RESPONSE --> Both of these triangles have vertical angles and two sides equal.
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14:18:39 SAS: Angle AOB and Angle FOE are equal because they are vertical angles, so we have 2 sides and the included angle of triangle AOB equal, respectively, to 2 sides and the included angle of triangle FOE. Thus, the Side-Angle-Side property holds that triangle AOB is congruent to triangle FOE.
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RESPONSE --> Yes
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14:30:02 **** Explain the argument you used to show that the triangles were congruent.
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RESPONSE --> I used the properties of side to angle to side
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14:31:37 **** query 9.4.18 ACB and QPR similar triangles, C and P rt angles, A=42 deg **** List the measures of the three angles of each triangle and explain how you obtained each.
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RESPONSE --> a=42 c=90 180-42-90= 48 b=48 p=90 angle q is the same as a thus r is the same as b, or 48
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14:31:42 It is given that Angle A = 42 deg. and Angle C = 90 deg. Since all three angles must add up to equal 180 then Angle B = 48 deg. In the second triangle, Angle P must equal 90 deg. since it is a right angle. To find Angle R, 90(48) = 90R sp 4320 = 90R and 48 = R Angle R = 48 deg. To find Angle Q, 90/90 = Q/42 Q = 42 Angle Q = 42 deg.
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RESPONSE --> yes
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14:33:36 **** query 9.4.24 similar triangles, corresp sides a, b, 75; 10, 20, 25 **** What are the lengths of sides a and b and how did you obtain each?
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RESPONSE --> For A I used the following... 75*(10) = 25A 750 = 25A (750/25) A= 30 as for B I used... 75/25 = B/20 1500/25 = 25B /25 B = 60
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14:33:40 To find a, 75 (10) = 25a 750 = 25a a= 30 To find b, 75/25 = b/20 1500/25 = 25b/25 so b = 60. a = 30, b = 60 and c = 75. These values are triple the values of the similar triangle.
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RESPONSE --> Yes
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14:35:02 **** query 9.4.42 rt triangle a = 7, c = 25, find b **** What is the length of side b and how did you obtain it?
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RESPONSE --> For this I used the pythagorean theorom... 49 + b^2 = 625 49-49+b^2= 625-49 b^2 = 576 square rooted, b=24
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14:35:34 By the Pythagorean Theorem a^2 + b^2 = c^2. So we have 49 + b^2 = 625 Subtract 49 from both sides to get b^2 = 576. Take the square root of both sides to get b = 24.
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RESPONSE --> It says that you can take the leg lengths,squared, and get the square of the hypotenuse.
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14:35:37 **** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b?
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RESPONSE --> Ywes
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14:35:41 Student Response: It says the sum of the squares of the lengths of the legs is equal to the square of the hypotenuse. I showed that this is true in the previous problem. I squared the legs and they equaled the hyppotenuse squared.
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RESPONSE --> t
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14:37:25 **** query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5?
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RESPONSE --> (m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26/2 = 13 (m^2 - 1) / 2 = (5^2 - 1 ) /2 = 24/2 = 12 taking this into consideration with m=5 then the three parts of the pyhagorean is 12, 13, and 5
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14:37:35 ** If m = 5 then (m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26 / 2 = 13 (m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12 So the Pythagorean triple is 5, 12, 13. We can verify this: 5^2 + 12^2 should equal 13^2. 5^2 + 12^2 = 25 + 144 = 169. 13^2 = 169. The two expressions are equal so this is indeed a Pythagorean triple. **
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RESPONSE --> yes, exactly as I worked it
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14:37:59 **** How did you verify that your result is indeed a Pythagorean Triple?
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RESPONSE --> by running the answers through the pythag. thoereom.
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14:38:03 Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
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RESPONSE --> yes
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14:42:41 **** query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem. **** How high is the break, and how
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RESPONSE --> taking that the break is equal to b, the hypoten. is equal to 10ft. -b As the triangle iscreated from the standing section, the base(ground?), and the broken piece, the figures that we will be using are b, 10-b, and 3. b^2 + 3^2 = (10-b)^2 b^2+9=100-20b+b^2 9=100-20b -20b=-91 b=4.55 as such, we find that the break section is at 4.55 feet and thus the broken piece is 5.45 (10-4.55) long.
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14:43:10 did you obtain your result?
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RESPONSE --> By using the methods above as done with the pyth. theorom
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14:43:14 ** If the break is at height x then the hypotenuse, consisting of the broken part, is at height 10 - x. The triangle formed by the vertical side, the break and the ground therefore has legs x and 3 and hypotenuse 10-x. So we have x^2 + 3^2 = (10-x)^2. Squaring the 3 and the right-hand side: x^2 + 9 = 100 - 20 x + x^2. Subtracting x^2 from both sides 9 = 100 - 20 x so that -20 x = -91 and x = 4.55. The break occurs at height 4.55 ft and the broken part has length 10 - 4.55 = 5.45, or 5.45 feet. **
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RESPONSE --> yest
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14:43:46 **** How did the Pythagorean Theorem allow you to solve this problem?
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RESPONSE --> by using substitution of numbers
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14:43:50 I substituted the numbers into the Pythagorean Theorem.
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RESPONSE --> yes
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14:44:20 **** query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find
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RESPONSE --> I had a bit of trouble with this one, not sure why, but I don't like my results... as such I wish to see your answer
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14:44:24 it?
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RESPONSE --> eh>
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14:44:45 ** This problem is algebraically demanding. Your text might have a slicker way to do this, but the following works: If the equal sides are x then the base is 128 - 2 x. The altitude forms a right triangle with half the base and one of the equal sides. The sides of this right triangle are therefore 48, 1/2 (128 - 2x) = 64 - x, and x. The right angle is formed between base and altitude so x is the hypotenuse. We therefore have 48^2 + (64 - x)^2 = x^2 so that 48^2 + (64 - x) ( 64 - x) = x^2 or 48^2 + 64 ( 64-x) - x(64 - x) = x^2 or 48^2 + 64^2 - 64 x - 64 x + x^2 = x^2 or 48^2 + 64^2 - 128 x + x^2 = x^2. Subtracting x^2 from both sides we get 48^2 + 64^2 - 128 x = 0. Adding 128 x to both sides we get 48^2 + 64^2 = 128 x. Multiplying both sides by 1/128 get have (48^2 + 64^2) / 128 = x. Evaluating this expression we end up with x = 50. The base of the triangle is therefore 128 - 2x = 128 - 2 * 50 = 128 - 100 = 28. So its area is 1/2 b h = 1/2 * 28 * 48 = 672. ** DRV
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RESPONSE --> Ahh, hah... okay, I was wondering why it was getting so indepth
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14:44:57 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> not alot of surprises
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14:45:01 I had some trouble with the last 2 problems.
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RESPONSE --> yes
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Αƌbyd assignment #024 {~ήǪSi`患 Liberal Arts Mathematics II 05-15-2006 I܅oҝ±ؓJ assignment #024 {~ήǪSi`患 Liberal Arts Mathematics II 05-15-2006
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23:48:44 **** 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 --> As the sphere has a diameter of 14.8 I used the formula for finding a sphere's volume.. the radius is 1/2 the diameter of 14.8.... or 7.4 volume= 4/3*pi*7.4*7.4*7.4= 1696.4954
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23:48:55 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 --> Okay, my calculations were a wee bit off
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23:52:11 **** 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 --> I used the pyramid volume formula... v=1/3BH v=1/3(12x4)x10 v=.333x480 v=159.84
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23:52:29 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 --> Okay, if I had rounded off I would have gotten the correct result.
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23:53:05 **** 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 --> I'm confused, do we count just the cylinder part or the neck too?
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23:53:16 ** 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 --> Okay, that answered my question then
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23:55:16 **** 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 --> I had trouble with this one...
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23:55:24 ** 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 --> Okay, I think I understand now
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23:56:51 **** 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 --> Again, I seem to have trouble with this particular chapter....
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23:57:05 ** 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 --> I'm still a bit confused so I will go back and restudy this one
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23:57:36 **** 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 --> This one has given me trouble as well
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23:57:42 ** 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 --> I need to study this more
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23:58:05 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I had alot more trouble with this particular section... I need to go back and study section 9.5 in greater detail.
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23:58:10 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 --> yes
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坞gںT۠_͊ assignment #025 {~ήǪSi`患 Liberal Arts Mathematics II 05-15-2006
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00:00:53 **** query 9.7.6 intersecting lines m, n parallel to k **** In which geometry or geometries is this possible and why?
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RESPONSE --> This can happen if done on a sphere... somewhat like a globe i guess you would say... Riemannian geometery covers this
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00:00:59 ** Two parallel lines intersect on a sphere (think of lines of longitude). So this occurs in a Riemannian geometry. **
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RESPONSE --> yes
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00:01:42 **** query 9.7.18 ruler r.b. CD wrench nail **** To which of the objects is the coin topologically equivalent and why?
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RESPONSE --> The coin and the ruler are topologically the same as one another... also as the nail.... the reason is that none of these three have any form of holes.
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00:01:46 The coin is topologically equivalent to the ruler and the wrench nail because none of these have holes.
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RESPONSE --> yes
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00:02:04 **** query 9.7.27 genus of 3-hole-punched sheet of paper **** What is the genus of the sheet and why?
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RESPONSE --> As the paper has 3 holes the genus is 3
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00:02:38 **** query 9.7.42 3,3,3,3,4,4,2,2 **** Can the network be traversed or not and why?
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RESPONSE --> No, as this network has 4 odd vertices it cannot be traversed
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00:02:41 ** This network contains 4 odd vertices. A network with 0 or 2 odd vertices can be traversed; a network with 4 odd vertices cannot. **
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RESPONSE --> yes
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00:03:50 **** If you start on a vertex of order 3 can you traverse the network and end up on that vertex? Explain why your answer must be true.
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RESPONSE --> If you start on a vertex of 3 there is no way of ending up on that same three edges.... this is because you end your point on the vertex at the first edge.
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00:03:53 ** You cant start on a vertex of order 3 and end up on the same one. You leave the vertex along the first of the three edges. When you traverse the second of these edges you are returning to the vertex, and when you leave again you have to travel along the third and you can't get back. You can end up on a different vertex of degree 3 if there is one (and if there is one you must end on it), but you can't end up on the degree-3 vertex you started from. **
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RESPONSE --> yes
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00:07:47 **** If you start on a vertex of order 4 can you traverse the network and end up not on that vertex? Explain why your answer must be true.
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RESPONSE --> No.... if you start on order 4 you can leave on the 1st and return on the second but you wind up leaving again on the 3rd corner.
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00:07:52 ** If you start on a vertex of order 4 you cannot traverse the network without ending up on that vertex, since you leave the vertex on the first edge, return on the second and leave on the third. If you traverse the network you have to return to the vertex on the fourth edge, and you cant leave again. **
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RESPONSE --> Yes
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00:08:42 **** If you start on a vertex of order 2 and traverse the network must you end up on that vertex? Explain why your answer must be true.
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RESPONSE --> Yes... I guess if that's what you mean... because if you start on the order of 2 then you end up on 2
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00:08:49 ** If you start on a vertex of order 2 and traverse the network you leave on the first edge, return on the 2 nd and youre stuck there. **
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RESPONSE --> Yes, more or less
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00:08:55 No, because once again this is an even vertex. One point must be the starting point and one the ending point.
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RESPONSE --> Oh, okay....
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00:09:21 **** If you start off of a certain vertex of order 3 and traverse the network is it possible to end up somewhere besides this vertex?
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RESPONSE --> No, because you leave the vertex before you can return to these areas...
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00:09:24 ** If you start on a vertex of order 3 and traverse the network you leave on the first, return on the second and leave on the third edge. You cant travel any of these edges again so you can never return. Therefore you must end up elsewhere. **
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RESPONSE --> yes
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00:09:35 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This one was alot easier for me than the last.
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"