#$&* course Mth 151 8/30 3pm 012. The common sense of logic.
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Given Solution: I said what would happen under a certain condition. In situation #2, that condition is fulfilled and what I said would happen doesn't happen. Therefore in situation #2 it is clear that I wasn't telling the truth. In situation #3, the condition that I addressed isn't fulfilled so no matter what happens I can't be accused of not telling the truth. I said what would happen if rains. No matter what happens, if it doesn't rain what I said cannot be held against me. It should be clear to anybody that situation #1 is exactly what you would expect, and that situation #4 is just would you would probably expect from my statement in the event that it doesn't rain, so nobody would say that this situation violates my claim to truthfulness. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I missed the #3 statement because of not reading the question clearly ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q002. Suppose that tell you 'It will rain today and I will give you $100'. Under which of the following circumstances can you claim that I was not telling the truth? 1. It rains and I give you $100. 2. It rains and I don't give you $100. 3. It doesn't rain and I give you $100. 4. I doesn't rain and I don't give you $100. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It will rain today and I will give you $100 for you to be telling the truth both conditions must be met 1. It rains and I give you $100. True both conditions are met 2. It rains and I don't give you $100. False Only one condition is met 3. It doesn't rain and I give you $100. False Only one condition is met 4. It doesn't rain and I don't give you $100. False Only one condition is met confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: It should be clear that situation #1 completely fulfills the conditions of my statement. Both of the things that I say will happen do happen. In situation #2, it rains but you don't get the $100. I said two things were going to happen and one of them didn't. In that case you would have to say that I wasn't telling truth. In situation #3, again one of the things I say is going to happen does but the other doesn't, so again you would have to say that I wasn't telling truth. In situation #4, neither of the things I say will happen does and certainly it would have to be said that I wasn't telling truth. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q003. Suppose that tell you 'It will rain today or I will give you $100, but not both'. Under which of the following circumstances can you claim that I was not telling the truth? 1. It rains and I give you $100. 2. It rains and I don't give you $100. 3. It doesn't rain and I give you $100. 4. I doesn't rain and I don't give you $100. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'It will rain today or I will give you $100, but not both' 1. It rains and I give you $100. False you do both 2. It rains and I don't give you $100. True only one is done 3. It doesn't rain and I give you $100. True only one is done 4. It doesn't rain and I don't give you $100. False both are done confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In situations 2 and 3, one of the things happens and the other doesn't, so you would not be able to say that I wasn't telling the truth. However in situation 1, both things happen, which I said wouldn't be the case; and in situation 4 neither thing happens. In both of these situations you would have to say that I was not telling truth. STUDENT COMMENT: :im not sure if I don’t understand or if there is just a typo present it would seem to me that 2 and 3 show that one situation was fulfilled and not the other which is what you said would happen INSTRUCTOR RESPONSE: In situation 1, I both things happen. However my statement include the phrase 'but not both'. In situation 4, neither thing happens. My statement said that one would happen. So in both situations 1 and 4, my statement fails to be true. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q004. Suppose that tell you 'It will rain today or I will give you $100'. Under which of the following circumstances can you claim that I was not telling the truth? 1. It rains and I give you $100. 2. It rains and I don't give you $100. 3. It doesn't rain and I give you $100. 4. It doesn't rain and I don't give you $100. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It will rain today or I will give you $100 Will be true as long as at least one criteria is fulfilled 1. It rains and I give you $100. True met both and it only had to meet one 2. It rains and I don't give you $100. True met one 3. It doesn't rain and I give you $100. True met one 4. It doesn't rain and I don't give you $100. False did not meet either criteria confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: At first this might seem to be the same as the preceding problem. But in the preceding problem we specifically said '... but not both.' In this case that qualification was not made. Therefore we have regard the statement as true as long as at least one of the conditions is fulfilled. This is certainly the case for situation 1: both conditions are true we can certainly say that at least one is true. So in situation #1 we have to regard the present statement as true. So situation #1 would not be included among those in which I could be accused of not telling the truth. STUDENT COMMENT: Im not sure weather 2 was included in the statement that were not true INSTRUCTOR RESPONSE: In situation 2 one of the conditions was fulfilled, satisfying the original 'or' statement. So situation 2 doesn't contradict the original statement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I found this exercise very easy. ------------------------------------------------ Self-critique rating #$&* 001. Areas ********************************************* Question: `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Area =length * width We will say the l=4m and w=3m A = 4m*3m A=12m^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2. FREQUENT STUDENT ERRORS The following are the most common erroneous responses to this question: 4 * 3 = 12 4 * 3 = 12 meters INSTRUCTOR EXPLANATION OF ERRORS Both of these solutions do indicate that we multiply 4 by 3, as is appropriate. However consider the following: 4 * 3 = 12. 4 * 3 does not equal 12 meters. 4 * 3 meters would equal 12 meters, as would 4 meters * 3. However the correct result is 4 meters * 3 meters, which is not 12 meters but 12 meters^2, as shown in the given solution. To get the area you multiply the quantities 4 meters and 3 meters, not the numbers 4 and 3. And the result is 12 meters^2, not 12 meters, and not just the number 12. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Easy pie ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A=½ * b * h A=½ *4m*3m A=½ *12m^2 A=6m^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h. STUDENT QUESTION Looking at your solution I think I am a bit rusty on finding the area of triangles. Could you give me a little more details on how you got your answer? INSTRUCTOR RESPONSE As explained, a right triangle is half of a rectangle. There are two ways to put two right triangles together, joining them along the hypotenuse. One of these ways gives you a rectangle. The common hypotenuse thus forms a diagonal line across the rectangle. The area of either triangle is half the area of this rectangle. If this isn't clear, take a blade or a pair of scissors and cut a rectangle out of a piece of paper. Make sure the length of the rectangle is clearly greater than its width. Then cut your rectangle along a diagonal, to form two right triangles. Now join the triangles together along the hypotenuse. They will either form a rectangle or they won't. Either way, flip one of your triangles over and again join them along the hypotenuse. You will have joined the triangles along a common hypotenuse, in two different ways. If you got a rectangle the first time, you won't have one now. And if you have a rectangle now, you didn't have one the first time. It should be clear that the two triangles have equal areas (allowing for a little difference because we can't really cut them with complete accuracy). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just had to remember the formula for area of a triangle ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Find the formula for area of a parallelogram A=b*h A=5m*2m A=10m^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A=1/2*5cm*2 A=1/2*10cm^2 A=5cm^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aIt is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self explainitory ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A= base * average altitude A=4km*5km A=20km^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aAny trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2. STUDENT SOLUTION ILLUSTRATING NEED TO USE UNITS IN ALL STEPS A=Base time average altitude therefore………A=4 *5= 20 km ^2 INSTRUCTOR COMMENT A = (4 km) * (5 km) = 20 km^2. Use the units at every step. km * km = km^2, and this is why the answer comes out in km^2. Try to show the units and how they work out in every step of the solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Find the average altitude 3 cm + 8cm = 11cm 11cm/2=5.5cm A=4cm*5.5cm A= 22cm^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just had to find the average altitude before I was able to find the area ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q007. What is the area of a circle whose radius is 3.00 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First thing you need to know is what the formula is for the area of a circle A= pi * rcm^2 A =3.14 * 9cm^2 A = 28.26 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThe area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought I knew what is was doing and I was correct ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Know your formula for the circumference of a circle C=2*pi*r C=2*3.14*3cm C=6.28*3cm C=18.84cm confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThe circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.8 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am on the right track, but why did you drop the .04 off the end of the answer.
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Given Solution: `aThe area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q010. What is the area of a circle whose circumference is 14 `pi meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I have to find the radius C=14* pi m r=14/2 r=7m a=pi*r^2 a=3.14*49m^2 a=153.86m^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2. STUDENT QUESTION: Is the answer not 153.86 because you have multiply 49 and pi???? INSTRUCTOR RESPONSE 49 pi is exact and easier to connect to radius 7 (i.e., 49 is clearly the square of 7) than the number 153.86 (you can't look at that number and see any connection at all to 7). You can't express the exact result with a decimal. If the radius is considered exact, then only 49 pi is an acceptable solution. If the radius is considered to be approximate to some degree, then it's perfectly valid to express the result in decimal form, to an appropriate number of significant figures. 153.86 is a fairly accurate approximation. However it's not as accurate as it might seem, since you used only 3 significant figures in your approximation of pi (you used 3.14). The first three figures in your answer are therefore significant (though you need to round); the .86 in your answer is pretty much meaningless. If you round the result to 154 then the figures in your answer are significant and meaningful. Note that a more accurate approximation (though still just an approximation) to 49 pi is 153.93804. An approximation to 5 significant figures is 153.94, not 153.86. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see then instead of multiplying the r^2 by pi physically just show you answer as followed 49m^2*pi ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q011. What is the radius of circle whose area is 78 square meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I will have to work backwards to find the radius C=78m^2 C=sqrt(78m^2/pi) C=sqrt(78/pi)m r=about 5m confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aKnowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m. STUDENT QUESTION Why after all the squaring and dividing is the final product just meters and not meters squared???? INSTRUCTOR RESPONSE It's just the algebra of the units. sqrt( 78 m^2 / pi) = sqrt(78) * sqrt(m^2) / sqrt(pi). The sqrt(78) / sqrt(pi) comes out about 5. The sqrt(m^2) comes out m. This is a good thing, since radius is measured in meters and not square meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just work backwards ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q012. Summary Question 1: How do we visualize the area of a rectangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t know really what you are asking in this question But Im going to give it my best shot The area of a rectangle is how much stuff will fit in the rectangle which is figured out by using the formula A=l*w confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t think of it as being filled with one unit squares but I think I have the jest of it ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q013. Summary Question 2: How do we visualize the area of a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A triangle is pretty much the same as the rectangle but instead of just being A=l*w You also have to multiply it by ½. A=1/2*l*w confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have never thought of doing it that way. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Chop one of the ends off and put it together just like a puzzle then it is in the shape of a rectangle A=l*w confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we will get the base And then find the average altitude Then we can set up the formula A=b*average altitude confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t think that orienting the trapezoid was necessary because it thought it was a given ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q016. Summary Question 5: How do we calculate the area of a circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the radius is given then solve A=pi*r^2 But if the diameter is given then you would have to divide it by two in order to get the radius r=d/2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe use the formula A = pi r^2, where r is the radius of the circle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for the circumference of a circle is C=2*pi*r I find it easy to avoid confusing it was area of a circle because it does not contain the exponent ^2 and is not measured in units ^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I like how yours reads compared to mine ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t really understand what you mean by organized you knowledge of the principles But I will try to answer the best I can A rectangle and a parallelograms areas are found with the same formula A=b*h A triangle is about the same but you have to multiply it by ½ so you would have A=1/2*b*h Area of a trapezoid is easy to remember because it is the most different other than a circle A=average altitude * b Area of a circle is just something you have to remember A=pi*r^2 But not to be confused with circumference of a circle C=2*pi*r ------------------------------------------------ Self-critique rating #$&* "