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
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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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