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course Phy 241
06/06/2013 11:34 am
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: The straight line connected from (7, 17) to the point (10, 29) is steeper than the other line because the slope of the line between (7, 17) to the point (10, 29) is 4, an unit more than the line between points (3, 5) and (7, 17).
confidence rating #$&*: 3
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Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5.Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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Your solution: The two values of x can make the expression zero because when either of the two interacts with the equations in the parentheses, one of the numbers will be a zero; in which in turn when a number is multiplied by 0 it will equal 0.
a) x = 2 b) x = -2.5
(2 - 2)*(2(2)+5) (-2.5 - 2)*(2(-2.5) + 5)
= (0)*(4+5) = (-4.5)*(-5 +5)
= (0)*(9) = (-4.5)*(0)
= 0 = (0)
confidence rating #$&*: 3
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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: The x values that will give zero are -4, -2, and 2.
x = -2 x = 2 x = - 4
(3(-2) - 6)*(-2+ 4)*((-2)^2 - 4) (3(2) - 6)*(2+ 4)*(( 2)^2 - 4) (3(-4) - 6)*(-4+ 4)*((-4)^2 - 4)
= (-12)*(2)*(0) = (0)*(6)*(0) = (6)*(0)*(64)
= (0) = (0) = (0)
confidence rating #$&*: 3
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Question: `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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Your solution: The second trapezoid has the greater area than the first trapezoid because the second trapezoid is 40 units wide, while the first trapezoid is 4 units wide.
confidence rating #$&*: 2
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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:
1. The graph of y = x^2 increases as the slope increases from left to right.
2. The graph of y = 1/x decreases as the slope decreases from left to right.
3. The graph of y = sqrt(x) increases as its slope increases from left to right.
confidence rating #$&*: 3
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Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
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Your solution: The frogs that I end up to have within each of the first three month are: 1st month = 22 frogs, 2nd month = 24.2 frogs, and the 3rd month = 26.62 frogs.
1st month 2nd month 3rd mouth
(20 * .1) + 20 (22 * .1) + 22 (24.2 * .1) + 24.2
= (2) + 20 = (2.2) + 22 = (2.42) + 24.2
= 22 = 24.2 = 26.62
In order to calculate the amount of frogs within the population is the equation of (20 * 1.1^300), with 300 representing the amount of months, and 1.1 being the equivalent of 10%.
confidence rating #$&*: 3
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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: The pattern from the equation, 1/x, is that the answer increase 10 times from the previous answers. The values of 1/x will increase without bound. If x continue to approach 0 through the values .1, .01, .001, .0001, ..., there is no restriction to how big the reciprocals can become.
The graph of y = 1/x becomes steeper and steeper as it gets close to the y axis, continuing to do so without bound but never touching the y axis.
confidence rating #$&*: 2
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Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
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Your solution: t=5, v = 3 t + 9 = (3*5) + 9 = 24;
E = 800 * 24^2
= 800 * 576
= 460800
confidence rating #$&*: 3
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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 = 800 v^2; v = (3t + 9)
E = 800 (3t + 9) ^2.
E = 800 (3t + 9) (3t + 9)
E = 800 (9 t^2 + 54 t + 81)
E = 7200 t^2 + 43320 t + 64800
confidence rating #$&*: 3
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Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero?
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Your solution: The x-values are the values of the expression of zero are 0, 5, and -3.
confidence rating #$&*: 3
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Question: One straight line segment connects the points (3,5) and (7,9) while another connects the points (3, 10) and (7, 6). 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?
Any solution is good, but a solution that follows from a good argument that doesn't actually calculate the areas of the two trapezoids is better.
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Your solution: The second trapezoid has the greater area than the first one since the second trapezoid is 2 units taller than the first trapezoid.
confidence rating #$&*: 3
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Question:
Suppose you invest $1000 and, at the end of any given year, 10% is added to the amount. How much would you have after 1, 2
and 3 years?
What is an expression for the amount you would have after 40 years (give an expression that could easily be evaluated using a calculator, but don't bother to actually evaluate it)?
What is an expression for the amount you would have after t years?
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Your solution:
Year 1 Year 2 Year 3
$1000(.1) + 1000 $1100(.1) + $1100 $1210(.1) + $1210
= $1100 = $1210 = $1331
The expression for the amount after t years is 1000 * 1.1^t.
confidence rating #$&*: 3
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