course Mth 164 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.
......!!!!!!!!...................................
RESPONSE --> We construct two columns and label the first column x and the second y . Put the numbers -3 -2 -1 0 1 2 3 and the x column. Substitute these numbers in for x and we get get the following values --> -13, -10, -7, -4, -1, 2, 5 respectively. The intercepts are (0, -4) and (1.3, 0) confidence assessment: 3
.................................................
GFoкщLev assignment #003 003. PC1 questions qa initial problems 01-13-2008
......!!!!!!!!...................................
21:17:59 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
......!!!!!!!!...................................
RESPONSE --> The line connecting (7, 17) and (10, 29) is steeper for this line has a larger slope than does the line connecting (3, 5) to (7, 17) confidence assessment: 2
.................................................
yސbC딗ŦDԏ assignment #001 001. Areas qa areas volumes misc 01-13-2008
......!!!!!!!!...................................
21:22:23 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.
......!!!!!!!!...................................
RESPONSE --> A= l x w A= 4 x 3 A = 12 self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:24:15 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> A = .5 x length x width A = .5 x 4 x 3 A = 6 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:26:08 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.
......!!!!!!!!...................................
RESPONSE --> A = b x h A = 5 x 2 A = 10 self critique assessment: 1
.................................................
......!!!!!!!!...................................
21:27:19 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> A = .5 x b x h A = .5 x 5 x 2 A = 5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:30:34 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> A = .5 x b x h A = .5 x 5 x 4 A = 10 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:32:41 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> A = b x (sum of h) A = 4 x 11 A = 44 kmsquared confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:35:01 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> A = n x 3sq A = n x 9 A = approximatly 28.27cm confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:37:46 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> C = d x pi C = 2(3) x pi C = 6 x pi C = 18.849cm confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:40:07 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> A = pi x r^2 r = .5 x d --> r = .5 x 12m --> r = 6 A = pi x 6^2 A = pi x 36 A = 113.097m confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:47:40 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.
......!!!!!!!!...................................
RESPONSE --> A = pi x r^2 C = 2 pi r r = 14 pi m / (2pi) r = (28) x (pi/pi) m r = 7 x 1m r = 7m self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:51:11 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> r = sq root of A / pi r = sq root of 78 / pi r = 4.983 sq m confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:53:39 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> A = length x width Area is the space ""inside"" the rectangle. ? confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:57:09 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> We visualize the area of a right triangle in the same way as a rectangle then mulitply by .5 whereas a right triangle is half of a rectangle. confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:57:36 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> I have no clue. confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:58:29 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> It's been a while since i've done this so i will have to brush up on my skills confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:00:28 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> A = pi x r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:04:28 `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?
......!!!!!!!!...................................
RESPONSE --> C = d x pi whereas A = pi x r^2 confusion may be where is d does not equal r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:07:17 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> much of this knowledge is common knowledge that must be known to do many of the operations confidence assessment: 2