Precalculus

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course Mth 158

Submitted August 30, 2010 @8:37pm. - Precalculus

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?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When you plot these x-y coordinates (3,5) and (7,17) you can determine the slope by the rise over run(the distance you move up and out between the points) and with these sets of coordinates, the rise is 12 and the run is 4. On the second set of x-y coordinates, (7,17) and (10,29) the rise is 12 and the run is 3. So the second line has a slight steeper slope.

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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 have the expression equal zero, one of the two expressions within the parenthesis must equal zero. When you substitute 2 for x in the first expression within parentheses, the answer is zero, therefore whatever the product of the second expression is then multiplied by zero which equals zero. When you substitute -2.5 for x in the second expression within parentheses; 2*-2.5 = -5 and -5+5 = 0. Therefore, whatever the product of the first expression within parentheses will be multiplied by 0 which will equal 0.

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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 only values for x that will result in a product of zero is 2, -2, and -4 as one of the expressions within parentheses must have a product of zero for the entire expression to equal 0. Using 2 in the first expression will give a product of zero: 3(2) - 6 = 0. Using -4 in the second expression will also give a product of zero: -4+4 = 0. Using -2 in the third expression also gives a product of zero: -2^2 = 4 and 4-4= 0.

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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:

To find the area of these trapezoids you use the formula area = a[b1+b2]

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The altitude of the first trapezoid with the points (3,5) and (7,9) is 7-3 = 4 and the two bases are 5 and 9 so you add these together and divide by 2. 5+9=14/2 = 7*4 = 28. The altitude of the second trapezoid with the points (10,2) and (50,4) is 50-10=40 and the two bases are 2 and 4 so you add these together and divide by 2. 2+4=6/2= 3*40 = 120. Therefore, the second trapezoid has the greater area.

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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:

For the expression y=x^2 with x>0, I used x coordinates of 1,2,3 and 4 which gave me the y values of 1, 4, 9, and 16. The graph of these points show it increases as its slope increases.

For the expression y=1/x with x>0, I used the x coordinates of1,2,3, and 4 which gave me the y values of 1, ½, 1/3, and ¼. This graph decreases as its slop decreases.

For the expression y=sort(x) with x>0, I used the x coordinates of 1,2,3, and 4 which gave me the y values of 1, 1.41, 1.732, and 2. This graph increases as its slope decreases.

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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:

1 month = 20 frogs *10% = 2, 20+2 = 22

2 month = 22 frogs * 10% = 2.2, 22+2.2 = 24.2

3 month = 24.2 frogs * 10% = 2.42, 24.2 + 2.42 = 26.62

20 frogs * 10% * 300

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Self-critique (if necessary):

I really wasn’t sure where to start with this last question but after seeing your solution, I understand what I should have done.

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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:

When you calculate 1/x with the x values being 1, .1, .01, and .001 the solution to each calculation increase by 10 and there is no limit as to how far you could go. The graph gets steeper.

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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 = 3 t + 9

E = 800 v^2

t = 5

3(5) + 9 = 24 so V = 24

E = 800(24^2) = 460800

The energy of the automobile at clock time t=5 is 460800.

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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:

V =3(5) + 9 = 24 so E = 800(24^2) = 460800

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