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course MTH 158
Question: `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
Your solution: It has 3 rows that has 4 squares in each row. The total area of the rectangle would be the rows times the square which would equal 12.
confidence rating #$&*: 2
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Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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Your solution: To find the right triangle length in meters you have to do the 4.0 *3.0= 12 meters. Then you would have to take half of that because of the ½ * b * h which would equal 6 meters.
Self Critique: The only thing that I do not understand Is why you have to use meters squared as the solution. I didn’t understand that unless its from where you take half of it where it is half a rectangle.
Self Critique Rating: 2
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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
Your solution: The first and only step is to times the two together because its P=b*h. So it would be 5*2 = 10m^2
Self Critique: I still don’t understand the meters squared and I had to look up the formula for a parallelogram.
Self Critique Rating: 2
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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
Your solution: It’s a ½ bh. So it would be ½ (5.0)(2.0) = The answer would be 5. You would times the 5 and 2 together giving you 10 and then you would take half of that which would be 5cm^2.
confidence rating #$&*: 3
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Self Critique: Im starting to get the reason why the cm^2 is there and why its used at the end of the answer. Im also getting better with knowing exactly what formula should be used on what problem.
Self Critique Rating: 3
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Question: `q005. Sketch on a set of x-y axes the four-sided quadrilateral whose corners are at the points (3, 0), (3, 7), (9, 11) and (9, 0) (just plot these points, then connect them in order with straight lines).
What would you say is the width of this figure, as measured from left to right?
If the width is measured from left to right, why does it make sense to say that the figure has 'altitudes' of 7 and 11?
Do you agree that the figure appears to be a quadrilateral 'sitting' on the x axis, with 'altitudes' of 7 and 11?
We will call this figure a 'graph trapezoid'. You might recall from geometry that a trapezoid has two parallel sides, and that its altitude is the distance between those sides. The parallel sides are its bases. There is a standard formula for the area of a trapezoid, in terms of its altitude and its two bases. We are not going to apply this formula to our 'graph trapezoid', for reasons you will understand later in the course.
The 'graph trapezoid' you have sketched appears to be 'sitting' on the x axis. An object typically sits on its base. So we will think of its base as the side that runs along the x axis, the side it is 'sitting' on.
The 'graph trapezoid' appears to be 'higher' on one side than on the other. We often use the word 'altitude' for height. This 'graph trapezoid' therefore will be said to have two 'graph altitudes', 7 and 11.
What therefore would you say is the 'average graph altitude' of this trapezoid?
If you constructed a rectangle whose width is the same as that of this trapezoid, and whose length is the 'average graph altitude' of the trapezoid, what would be its area?
Do you think this area is more or less than the area of the 'graph trapezoid'?
Your solution: The width is 6 points across. The reason why it has altitudes is because 7 and 11 are the highest points on the y intercept. The average graph altitude would be determined by finding the sum of the 7 and 11 by adding them together and dividing by 2 which gives you 9. You would times the average that I just found by the width which would give you the area which is 54. The area would be more I would believe because we’re only doing one half of the trapezoid.
Self Critique: I thought the area would be bigger but when I looked at the answer it said it would be equal I thought since you had a whole other side that you would have to equal in but it says you wouldn’t.
Self Critique Rating: 2
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Question: `q006. What is the area of a 'graph trapezoid' whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
Your solution: first you find the sum of the altitudes which would be 8+3 /2 which equals 5.5. Then you times that by the width which would be 4*5.5 = 22 cm^2.
Self Critique: I just used what I learned from the last question and applied it to this one.
Self Critique Rating: 2
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Question: `q007. What is the area of a circle whose radius is 3.00 cm?
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Your solution: A=pi*r^2 which then I would just fill the 3.00 cm in. which would be A=pi*(3.00)^2= 28.27
Self Critique: I was off .01 on the answer but that was from using the computer calculator probably I did have to look up the formula for the area of a circle.
Critique Rating: 2
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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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Your solution: 2*pi*r so all I have to do is fill in 3 for r which would be 2*pi*3 which would equal 18.9.
Self Critique: I think that im using the full pi instead of 3.14 so that’s why im off by decimal points.
Critique Rating: 2
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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?
Your solution: A= pi r^2 All I have to do is fill in for r with 6. You have to take have of the 12 to get r when doing diameter. The answer would be 113.09
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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?
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Your solution: The formula we would use is C/2 pi. Then we would just place the 14 pi in. So it would be 14pi/ 2pi which would equal 7 meters. Then you have to use A= pi*r^2 which would be pi *7^2 which equals 49*pi which equals 143.95.
Self Critique: I took one step to far and should have kept it at the 49*pi m^2. Im also wondering why I have to keep it at that point.
Self Critique Rating: 2
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Question: `q011. What is the radius of circle whose area is 78 square meters?
Your solution: First you use the formula Area=pi r^2 but were looking for r so we rearrange the formula to look like this r= sqrt(78m^2/pi)= 24.83
Self Critique: I do not understand how they got 5 and what steps I needed to take to get that answer.
Self Critique Rating: 2
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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?
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Your solution: We can visualize the rectangle had smaller squares that could in rows and columns. The formula for a rectangle is A= L*W so you would multiply the rows times the number of columns to get the area of a rectangle.
Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?
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Your solution: The area of a right triangles formula is a ½ *b*h. A right triangle is a rectangle cut in a half so you could vision the other half of the rectangle.
Self Critique: I didn’t mention anything about the altitude of the right triangle but the description was fine.
Self Critique Rating: 2
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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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Your solution: To calculate the area of a parallelogram you would just time its base and altitude together which would give you the area.
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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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Your solution: The only thing that you need to do is multiply the altitude by the width.
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Question: `q016. Summary Question 5: How do we calculate the area of a circle?
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Your solution: you would need the formula A=pi r^2 you would just plug in the number given and you will get an answer.
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Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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Your solution: you would use the formula C= 2pi r. In order to not get these to mixed up the c and the a helps me a stands for area and c stands for circumference. It might help just to memorize one of them as well.
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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
Your Response: I already knew most of these but I learned several tricks for the area of circles and some of the other equations as well.
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