course mth 163 assignment #003003. PC1 questions
......!!!!!!!!...................................
13:49:16 `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 way to find out which line is steeper is to figure out the slope for each line. We find the slope by getting the rise over the run. For the first line (3,5) and (7,17) we find the slope by: 17 - 5 = 12 and 7 - 3 = 4 12 / 4 = 3. The first line has a steepness of 3. The second line (7, 17) and (10,29) is: 29 - 17 = 12 and 10 - 7 = 3 12 / 3 = 4. The second line has a steepness of 4 and therefore is the steeper one. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:50:31 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
......!!!!!!!!...................................
RESPONSE --> I found the rise/run ratio to figure out which line was steeper. It is easier to figure it out this way, as oppose to drawing the lines out and trying to count the units to move. self critique assessment: 1
.................................................
......!!!!!!!!...................................
13:54:38 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
......!!!!!!!!...................................
RESPONSE --> If we substitute the x = 2 and x = -2.5 values into the expression we do find that both will equal zero. These can be the only two values to make the expression equal zero, because the values make one side of the expression equal zero and anything multiplied by zero will equal zero. (2 - 2) * (2*2 + 5) = 0 * 9 = 0 (-2.5 -2) * ( 2*-2.5 +5) = -4.5 * (-5 + 5) = -4.5 * 0 =0 confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:58:44 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
14:06:25 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
......!!!!!!!!...................................
RESPONSE --> The way to find what values x can be to make the expression (3x - 6 )*(x + 4) * (x^2 - 4) equal to zero, we just need to find what numbers will make one part of the expression equal zero. X can be 2: (3*2 - 6)*(2+4)*(2^2 -4) = (6-6)*(6)*(4 - 4) We can see that when x = 2, both the (3x-6) and (x^2 -4) parts will be equal to zero. X can be -2: The (x^2 - 4) would be (-2^2 -4) = 4 -4 = 0 X can be -4; The (x + 4) would be (-4 + 4) = 0 confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:08:48 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
......!!!!!!!!...................................
RESPONSE --> That's the same way I went about it, if you can figure out for each part of the expression what value would make it be equal to zero. self critique assessment: 1
.................................................
......!!!!!!!!...................................
14:21:21 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
......!!!!!!!!...................................
RESPONSE --> Just from looking at the two trapezoids on the graphs, it would appear that the trapezoid made with the (10,2) and (50,4) would have the greater area. We think about it this way, the trapezoid with (3,5) and (7,9). It only covers the x axis for 4 units and the highest point is 9 units up the y axis. The (10,2) and (50,4) trapezoid covers an area of 40 units on the x axis but only has a high point of 4 units on the y axis. Just from that information there we could say the (10,2) and (50,4) has the greater area since it covers 40 units. confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:22:27 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
......!!!!!!!!...................................
RESPONSE --> I didn't think of finding the average altitude for both the trapezoids. self critique assessment: 1
.................................................
......!!!!!!!!...................................
14:38:31 * `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
......!!!!!!!!...................................
RESPONSE --> For the graph y =x^2, it would best fit the description: as we move from left to right the graph increases as its slope increases. The line gets higher rapidly on the graph. For the graph y = 1/x we would say: as we move from left to right the graph decreases as its slope decreases. The line starts out at 1 on the y axis and goes down more for every unit moved over on the x axis. For the graph y = 'sqrt(x) we would say: as we move from left to right the graph increases as its slope decreases. The line does not climb up rapidly, yet it climbs the y axis slower as it moves over units on the x axis. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:43:52 For x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
......!!!!!!!!...................................
RESPONSE --> I cannot remember fore sure what I said about the y=1/x graph. The negative numbers approaching zero would be considered increasing. self critique assessment: 1
.................................................
......!!!!!!!!...................................
15:13:49 `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
......!!!!!!!!...................................
RESPONSE --> I know that I've done a problem like this before but I cannot think of how you can get the correct formula. confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:18:49 At the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
......!!!!!!!!...................................
RESPONSE --> I knew how to get the total number of frogs for the first 3 months but couldn't figure out how to find the total number of frogs for 300 months. I see how it would work to do 20 * 1.1^300. self critique assessment: 1
.................................................
¡y¢KՖ¬á’xiàÖ˜å‡È²»›U¤þIŸ assignment #003 003. PC1 questions qa initial problems 06-10-2008
......!!!!!!!!...................................
16:07:22 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
......!!!!!!!!...................................
RESPONSE --> First we'll substitute the values in for x: x = 1, 1/1 = 1 x = .1, 1/.1 = 10 x = .01, 1/.01 = 100 x = .001, 1/.001 = 1000 The answers correspond with the values for x, when x equals one tenth, 1/.1 equals 10, when x equals one hundrendth, 1/.01 equals 100 and so on. The values of x are approaching zero, one tenth, one hundrendth, one thousandth gets closer to zero each time. If we kept on approaching zero with the x value we would come up with the same pattern for the answers. We could use .0001, keep adding a zero after the decimal. We'd keep getting an extra zero added onto the answer, 1/.0001 = 10,000. The graph would be decreasing, would ""look"" like it touches the the x axis, but it wouldn't. The graph would get very close, but never touch. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:09:55 If x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph .
......!!!!!!!!...................................
RESPONSE --> The values would be without bound for the values of 1/x. We could keep going on forever and ever, never getting to zero. I was wrong on describing my graph, the line would get steeper and never touch with y axis. I was thinking of it wrong. self critique assessment: 1
.................................................
......!!!!!!!!...................................
16:13:40 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
......!!!!!!!!...................................
RESPONSE --> If v = 3 t + 9 and we know that t = 5 we can find; v = 3*5 + 9 = 15 + 9 = 24 Now that we found v = 24, we can find the energy since E = 800 v^2: E = 800 *24^2 = 800 * 576 = 460,800. Energy of motion is 460,800 self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:14:19 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
......!!!!!!!!...................................
RESPONSE --> I come to the same answer. All we have to do is substitute the values into each expression. self critique assessment: 1
.................................................
......!!!!!!!!...................................
16:15:57 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
......!!!!!!!!...................................
RESPONSE --> If v = 3 t + 9 and E = 800 v^2 then when could put the two expressions together and say: E = 800*(3t + 9)^2 This would gives us the same answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:17:17 Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................