course Mth 271
003. PC1 questions
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Question: `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?
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Your solution:
To find the slope of these lines you can use the equation:
(y1-y2)/ (x¬1-x2)
Line 1
((5) – (17))/((3) – (7))= (-12)/(-4) = 3/1
Line 2
((17)-(29))/((7) – (10)) = (-12)/(-3) = 4/1
Line two, 4/1, is steeper because rise is over run meaning the line would go up 4 and over 1 as opposed to line one, which would go up 3 and over 1.
Confidence Assessment:
I feel confident about this problem.
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Self-critique (if necessary):
OK
Self-critique Rating:
To find the slope of a line, you use two points and plot them in the equation (y1 – y2)/(x1 – x2). This is how you find rise over run. Simply plot the points and you find the slop.
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Question: `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.
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Your solution:
To prove that these are the only two solutions for x, the two equations should be set to zero and the value of x substituted in:
X - 2 = 0
(2) – 2 = 0
0 = 0
(2-2)(2x+5) = 0 this is true
2x+5 = 0
2(2.5) + 5 = 0
0 = 0
(x – 2)(2x+5) = 0 this is also true
((2)-2)(2(2.5)+5) = 0
(0)(0) = 0
Confidence Assessment:
I feel confident about this problem.
Self-critique (if necessary):
OK
Self-critique Rating:
By substituting the given values into the equation, I can prove that these are the only two values that can make the expression zero.
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Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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Your solution:
3x – 6 = 0
3x = 6
x = 2
x + 4 = 0
x = -4
x^2 – 4 = 0
x^2 = 4
x = +or- 2
x = 2, -4, -2
Confidence Assessment:
I feel very confident about this problem.
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Self-critique (if necessary):
OK
Self-critique Rating:
For this problem, I merely set each of the equations to zero and solved for x.
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Question: `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'.
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Your solution:
The coordinates of the first line, (3.5) and (7, 9), indicate that the line is 4 units wide and the coordinates of the second line, (10, 2) and (50, 4), indicating that this line is 50 units wide. For a square, the width of the square does not change, but the height does. To find the height, you need to average the coordinates of y. So average 5 and 9 to find 7 for the first line and for the second line, average 2 and 4, to get 3. To get the area we multiply the width by the altitude.
Line 1: 7 * 4 = 28
Line 2: 3 * 40 = 120, meaning that the trapezoid made by line two has a greater area.
Confidence Assessment:
OK
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Question: `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.
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Your solution:
To find the solution, I plugged in random numbers into the equation to find the trend of the graph. For the equation y = x^2, the value of y increases at a higher rate as the value of x increases.
For the equation y = 1/x, the value of y decreases as the value of x increases.
For the equation y = `sqrt(x), the value of y increases at a decreasing rate as x increases.
Confidence Assessment:
I feel fairly confident about this problem.
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Self-critique (if necessary):
OK
Self-critique Rating:
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Question: `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?
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Your solution:
To find what 10% of 20 is, you would multiply 20*.10 to find 2. To find how many frogs you would have at the end of the first month, you add 2 to 20 to get 22 total frogs. To find how many frogs you would have at the end of the second month you multiple 22 * .10 to get 2.2 and add that to 22 to get 24.2 total frogs. To get the total number of frogs at the end of month three, multiply 24.2*.10 to get 2.42 and then add that to 24.2 to get 26.62
Confidence Assessment:
OK
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Self-critique (if necessary):
OK
Self-critique Rating:
To find the total number of frogs, you have to find how many frogs will be added at the end of each month. The key to this problem is to find the number of additional frogs per month and not try to find the total for all three months at the same time.
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Question: `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?
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Your solution:
1/x
x = 1
1/1 = 1
X = .1
1/.1 = 10
X = .01
1/.01 = 100
X = .001
X = 1000
The smaller the denominator, the larger the number obtained. We say that the values of x are approaching zero because the values are getting smaller and getting closer to zero. to continue approaching zero, we could use .0001 and .00001 and so on. The values of 1/x get larger as the value of x gets smaller. The graph of y = 1/x has a steeper slope as the line approaches the y axis, but never touches the y axis.
Confidence Assessment:
I feel confident about this problem.
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Self-critique (if necessary):
OK
Self-critique Rating:
To find the solution to this problem, we enter the values into the given equation and evaluate what this means about the line.
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Question: `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?
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Your solution:
v = 3t + 9
E = 800v^2
v = 3(5) + 9
v = 15 + 9
v = 24
E = 800(24)^2
E = 460800
Confidence Assessment:
I feel extremely confident about this problem.
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Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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Your solution:
E = 800v^2
v = 3t + 9
E = 800(3t + 9)^2
Confidence Assessment:
Great!
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Self-critique (if necessary):
OK
Self-critique Rating:
To find the answer to this problem, I merely entered the value of v (which included t in the equation) into the equation for E.
&#Good responses. Let me know if you have questions. &#