course Mth 151 ȵM{vqassignment #001
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14:23:48 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). The evaluate each expression for x = 2.
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RESPONSE --> The brackets change the order of operations in the problem. In the second problem you must do the brackets first. 2-2/2+4...Division before addition and subtraction 2-1+4=5 [2-2]/[2+4]...Brackets first 0/6=0 confidence assessment: 3
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14:24:05 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> Ok. self critique assessment: 3
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14:26:14 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> The brackets determine to which power 2 is raised. 2^2+4...Raise 2 to the second power 4+4=8 2^[2+4]...Do brackets 2^6=64 confidence assessment: 3
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14:26:27 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> Ok. self critique assessment: 3
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14:30:27 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> The numerator is: 3 The denominator is: [2x-5^2*3x+1] 2-3 / [ [2(2)-5]^2*3(2)+1] -2+7(2)...Do brackets 2-3 / [7] -2+7(2)...Divide and multiply 2-(3/7)-2+14...Add and subtract 13.57 confidence assessment: 3
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14:32:00 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7. COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation? INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression. If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute. If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped. If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].
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RESPONSE --> Ok. self critique assessment: 3
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14:33:41 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> [4-5] ^ 2(4)-1 + 3 / 4-2...Do brackets [-1] ^ 2(4)-1 + 3 / 4-2...Do exponents 4-1+3/4-2...Do division 4-1+.75-2...Add and subtract 1.75 confidence assessment: 3
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14:34:42 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.
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RESPONSE --> Ok. self critique assessment: 3
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14:35:38 *&*& Standard mathematics notation is easier to see. On the other hand it's very important to understand order of operations, and students do get used to this way of doing it. You should of course write everything out in standard notation when you work it on paper. It is likely that you will at some point use a computer algebra system, and when you do you will have to enter expressions through a typewriter, so it is well worth the trouble to get used to this notation. Indicate your understanding of the necessity to understand this notation.
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RESPONSE --> I understand. self critique assessment: 3
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c攡Rݝ~߰ɚ assignment #001 001. Areas qa areas volumes misc 01-27-2007
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14:48:02 `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.
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RESPONSE --> A=L*W A=4*3 A=12m^2 confidence assessment: 3
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14:48:41 A 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.
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RESPONSE --> Ok. self critique assessment: 3
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14:53:54 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> A=.5*B*H A=.5*4*3 A=6m^2 confidence assessment: 3
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14:54:45 A 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.
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RESPONSE --> Ok. self critique assessment: 3
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14:59:34 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> A=B*H A=5*2 A=10m^2 confidence assessment: 3
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15:01:09 A 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:05:09 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> A=.5*B*H A=.5*5*2 A=5cm^2 confidence assessment: 3
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15:07:29 It 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:10:15 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> A=B*H A=4*5 A=20km^2 confidence assessment: 3
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15:10:52 Any 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:16:55 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> A=.5*H*(A+B) A=.5*4*(3+8) A=22cm^2 confidence assessment: 2
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15:18:20 The 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:24:47 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> A=PiR^2 A=3.14 * 3^2 A= 28.3cm^2 confidence assessment: 3
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15:25:53 The 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:33:49 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C=2PiR C=2 * 3.14 * 3 C=18.84cm confidence assessment: 3
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15:34:24 The 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 = 18.84 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:39:37 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> A=PiR^2 A=3.14 * 6^2 A=113.04m^2 confidence assessment: 3
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15:40:08 The 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.
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RESPONSE --> Ok. self critique assessment: 3
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15:51:02 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> R=C/(2Pi) R=14/(2Pi) R=7 A=PiR^2 A=3.14 * 7^2 A= 153.86m^2 confidence assessment: 2
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15:54:09 We 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.
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RESPONSE --> Ok. self critique assessment: 3
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16:04:10 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> A=PiR^2 R^2=A/Pi R=Square RT (A/Pi) R=5m confidence assessment: 3
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16:04:42 Knowing 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.
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RESPONSE --> Ok. self critique assessment: 3
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16:06:26 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> By being filled with one unit squares. confidence assessment: 3
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16:06:39 We 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.
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RESPONSE --> Ok. self critique assessment: 3
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16:07:42 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> By putting another right triangle and forming a square. confidence assessment: 3
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16:08:48 We 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.
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RESPONSE --> I said square instead of rectangle. self critique assessment: 2
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16:10:46 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Area is equal to base times height A=B*H confidence assessment: 3
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16:12:57 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> Ok. self critique assessment: 3
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16:18:33 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Area is equal to one half A plus B, times the height. A=.5(A+B)H confidence assessment: 3
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16:19:50 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> Ok. self critique assessment: 3
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16:22:16 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> Area is equal to pi times radius squared. A=PiR^2 confidence assessment: 3
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16:22:26 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> Ok. self critique assessment: 3
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16:26:13 `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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RESPONSE --> Circumference is equal to 2pi times the radius. C=2PiR When finding the area you have squared units which aren't found in circumference. confidence assessment: 3
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16:26:29 We 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.
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RESPONSE --> Ok. self critique assessment: 3
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16:28:50 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have refreshed my memory of area and circumference equations. confidence assessment: 3
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