course Mth 173 This ends the first assignment.TܩkI`y}Č
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12:13:55 `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 --> the area of the rectangle is 12 meters confidence assessment: 3
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12:15:10 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: 2
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12:16:38 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> the area of the right triangle is 6 confidence assessment: 0
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12:17:11 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: 2
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12:18:14 `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 --> the area is 15 confidence assessment: 2
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12:19:23 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 i see that the area is found with the equation A=b * h self critique assessment: 3
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12:20:27 `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 --> the area is 2.5 confidence assessment: 2
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12:21:34 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 --> the formula used in this equation is 1/2 * b *h self critique assessment: 2
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12:27:45 06-18-2007 12:27:45 `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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NOTES -------> the areais 25
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12:27:52 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:
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12:28:18 `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 --> the are is 24 confidence assessment: 0
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12:29:11 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 --> the formula is (b1 + b2)/ 2 self critique assessment: 3
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12:30:38 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> the are of the circle is 18.64 confidence assessment: 3
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12:31:43 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:
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12:35:41 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> the formula C=pi * D so the equation is C = 3.14 * 6 so the answer is 18.84 confidence assessment: 3
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12:35:59 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:
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12:37:45 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> the area is 37.68 confidence assessment: 3
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12:38:29 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 --> the area is A = pi r^2 self critique assessment: 3
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12:40:26 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> the area is 15.59 confidence assessment: 3
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12:40:39 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: 1
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12:41:26 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> the radius 4.5 confidence assessment: 2
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12:41:41 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: 2
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12:46:55 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> the area of a rectangle is viewed as rows and columns of identical squares. confidence assessment: 3
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12:47:09 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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12:48:34 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> the area of a triangle is A = (1/2) * b * h confidence assessment: 2
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12:48:52 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 --> ok self critique assessment: 2
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12:50:14 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> the are of a parallelogram is A = b * h confidence assessment: 3
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12:50:25 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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12:52:17 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> the are of a trapezoid is A = a( (b1 + b2)/2) confidence assessment: 3
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12:52:32 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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12:53:52 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> the area of a circle is calculated using the formula A = pi * r^2 confidence assessment: 3
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12:54:06 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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12:56:51 `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 --> the circumference of a circle is calculated by using the formula C = pi * d, diameter being diameter. because it has C in stead of A at the beginning of the equation. confidence assessment: 3
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12:57:29 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: 2
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12:57:52 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> i saved them as notes that i wrote down. confidence assessment: 3
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