precalculus

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course mth 163

1/13/12 10:05 p.m.

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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:

The first line has a slope of 3 because there are 12 changes in y for every 4 changes in x. The other line has a slope of 4 because there are 12 changes in y for every 3 changes in x. Therefore, the second graph has more of a slope and would be steeper than the first graph.

confidence rating #$&*:3

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Given Solution:

`aThe 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.

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Self-critique (if necessary):

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Self-critique Rating:OK

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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: If x =2, 2-2 would be 0 and everything that is multiplied by 0 is of course 0. This is also true for x= -2.5 which would make the second half of the expression 0 and thus make the entire expression 0 when multiplied. These are the only two numbers that could make either set of parentheses 0, and therefore are the only numbers that can make the expression 0.

confidence rating #$&*:3

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Given Solution:

`aIf 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.

STUDENT QUESTION

I think I have the basic understanding of how x=2 and x=-2.5 makes this equation 0

I was looking at the distributive law and I understand the basic distributive property as stated in algebra

a (b + c) = ab + ac and a (b-c) = ab - ac

but I don’t understand the way it is used here

(x-2)(2x+5)

x(2x+5) - 2(2x+5)

2x^2 + 5x - 4x - 10

2x^2 + x - 10.

Would you mind explaining the steps to me?

INSTRUCTOR RESPONSE

The distributive law of multiplication over addition states that

a (b + c) = ab + ac

and also that

(a + b) * c = a c + b c.

So the distributive law has two forms.

In terms of the second form it should be clear that, for example

(x - 2) * c = x * c - 2 * c.

Now if c = 2 x + 5 this reads

(x-2)(2x+5) = x * ( 2 x + 5) - 2 * (2 x + 5).

The rest should be obvious.

We could also have used the first form.

a ( b + c) = ab + ac so, letting a stand for (x - 2), we have

(x-2)(2x+5) = ( x - 2 ) * 2x + (x - 2) * 5.

This will ultimately give the same result as the previous. Either way we end up with 2 x^2 + x - 10.

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Self-critique (if necessary):

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Self-critique Rating:ok

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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

+6 +6

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3x=6

Divide by three on both sides to get x=2 which would make the first set of parentheses 0 and thus the expression 0 when multiplied.

X+4=0

-4 -4

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X= -4 therefore -4 will make the second set 0 and the entire expression 0.

x^2-4=0 is a difference of squares so I set x-4=0 and x+4 =0 which gave me x=2 and x= -2 which would make that parenthesis 0 and the entire expression 0.

confidence rating #$&*:3

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Given Solution:

`aIn 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.**

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Self-critique (if necessary):

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Self-critique Rating:ok

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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: When I attempted to graph this, I had trouble running out of lines on the paper but I improvised and was able to determine that the second trapezoid is much larger than the first due to the size of the images on the graph paper and the fact that there is significantly more distance between the x points 10 and 50 and the others at 3 and 7. While the points (3,5) and 7, 9 may be taller, the other trapezoid was much wider.

confidence rating #$&*:3

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Given Solution:

`aYour 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.

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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 graph of y=x^2, the solutions are, (1,1), (2,4), (3,9), (4,16). The solution is “As we move from left to right, the graph increases as its slope increases.”

For y=1/x, the solutions are (1,1), (2,.5), (3,.33), (4,.25). The description is “As we move from left to right the graph decreases as the slope increases”.

For y=’sqrt(x), the solutions are (1,1), (2, 1.414), (3,1.732), (4,2). The description would be “As we move from left to right the graph increases as the slope decreases”.

confidence rating #$&*:3

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Given Solution:

`aFor 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.

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Self-critique (if necessary):

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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: 20(.10)=2 so 20 +2 for the first month =22. Then 22(.10)=2.2+22=24.2 for the second month. So to figure out how many for 300 months you would have to ??

confidence rating #$&*:2

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Given Solution:

`aAt 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. We therefore get

20 * 1.1 = 22 frogs after the first month

22 * 1.1 = 24.2 after the second month

etc., multiplying by for 1.1 each month.

So after 300 months we will have multiplied by 1.1 a total of 300 times. This would give us 20 * 1.1^300, whatever that equals (a calculator, which is appropriate in this situation, 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.

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Self-critique (if necessary): I had difficulty figuring out how to project for 300 months from now. When viewing the solution, I see that I would have raised the 1.1 to the 300th power. I just wasn’t thinking about it in that way due to inexperience dealing with powers that are that high. I now understand the problem and solution.

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Self-critique 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: Each solution is like multiplying the previous solution by itself. You are doubling the previous solution to obtain the next one. The answers are 1, 10, 100, 1000. The values of x are approaching zero because we are moving from 1 downward to one tenth, then one hundredth, then one thousandth which are all smaller than 1 and as we continue we are getting closer and closer to zero. It would be a very large graph and without limit due to the amount of times you could multiply the numbers. It gets steeper as it goes because of the distance between the slopes getting greater and greater.

confidence rating #$&*:3

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Given Solution:

`aIf 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 .

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Self-critique (if necessary):I understand the question and answer but when reading the answer given, I had never heard of the term asymptote.

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Self-critique Rating:ok

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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 E=800 (3t+9) (3t+9) t=5 substitue the v= in for the v in the E= expression

E=800{3(5)+9}{3(5)+9} substitute the 5 in for the t in the equation

E=800(24)(24) multiply the 3 and 5 and add the 9 in each parentheses

E=460800 and finally multiply the 800 and 24 and 24 to get the answer.

confidence rating #$&*:3

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Given Solution:

`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 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: E=800(3t+9)^2

confidence rating #$&*:3

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Given Solution:

`aSince 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).

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