B3 - Precalculus

#$&*

course Mth 272

6/1 6pm

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 solve for slope: ( y2 - y1 ) / ( x2 - x1 ).

First equation: (17-5) / (7-3) = 12/4 = 3.

Second equation: (29-17) / (10-7) = 12/3 = 4.

Therefore the second equation is steeper because it does not stretch (run) as much as the first even though the rises are the

same.

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): OK

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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:

When x=2 in the expression (x-2) = 2-2 = 0 which when multiplied by any number results in zero.

When x=(-2.5) in the expression (2x+5) = 2(-2.5) + 5 = (-5) + 5 = 0 which makes any number zero when multiplied.

confidence rating #$&*: 3

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Given Solution: OK

`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): OK

I did not feel that the distributive process was necessary as the individual expressions are all that mattered, not them

together. But I do understand the process.

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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:

Set each expression to zero:

3x-6 = 0

3x = 6

x = 2

x+4 = 0

x = -4

x^2-4 = 0

x^2 = 4

x = 2 (take the square root)

When plugging in 2, (-4), or 2 for x, your answer will result in zero.

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): OK

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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:

Because the points of the second shape are so much farther apart (10 to 50 as compared to 3 to 7), the second shape must have a

larger area. The first shape is higher but the second is significantly wider.

Confidence 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 y = x^2 as you put in positive numbers the results get a lot bigger (4 comes out as 16), therefore the graph increases as

the slope increases.

For y = 1/x as you put in positive numbers the results get smaller and decrese by less every time (4 comes out as 0.25),

therefore the graph decreases as the slope decreases while getting closer to zero.

For y = sqrt(x) as you enter 1 you get 1, as you enter 2 you get 1.4, as you enter 3 you get 1.7 and 4 comes out as 2.

Therefore the graph increases with a decreasing rate.

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): OK

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Self-critique Rating: OK

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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 get the first month you will multiply 20 by 10% to get 2 then add that back to 20 to have 22 frogs by the end of the first

month.

For the second month you would multiply the new total, 22, by 10% to get 2.2 then add that back to 22 to have 24.2 frogs by the

end of the second month.

For the third month 24.2 * .10 = 2.42 then add back 24.2 + 2.42 = 26.62 frogs at the end of the third month.

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 did not do the 300 months and was unsure how to calculate the 300 months in advance. I now understand where the 1.1 came from

the original calulations and why you would raise it to the 300th power.

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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:

The values are approaching zero because the decimal keeps getting smaller but since the values are the divisor they go into 1

more resulting in 1, 10, 100, and 1000.

To continue approaching zero one might use .0001, .00001, etc. and the results will keep getting bigger like 10000 and 100000

respectively.

The slope gets steeper but never touches the y-axis.

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): OK

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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:

First you must use the velocity equation and plug in t: v = 3(5) + 9 = 15 + 9 = 24

Second plug your answer for v into the energy equation: E = 800 (24)^2 = 800 * 576 = 460800

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:

Since you use velocity in the energy problem you just need to fill in what v equals (using t) in the energy equation: 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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:

Solve each for zero:

2^x - 1 = 0

2^x = 1

x = 0

x^2 - 25 = 0

x^2 = 25

x = 5

2x + 6 = 0

2x = -6

x = -3

The x values that would equal zero would be 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 is larger because the point goes up 5 spots higher while cancelling out the one that is lower.

confidence rating #$&*: 2

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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:

One year: 1000 * .10 = 100 + 1000 = 1100

Two years: 1100 * .10 = 110 + 1100 = 1210

Three years: 1210 * .10 = 121 + 1210 = 1331

40 years: 1000 * 1.1^40

t years: 1000 * 1.1^40

confidence rating #$&*: 3

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Self-critique Rating: OK"

Self-critique (if necessary):

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Self-critique (if necessary):

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