#$&* course Mth 174 6/1 7 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
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These two expressions are being multiplied. This means that if either one of them equal 0 at an x-value, the whole expression itself must equal zero. The easiest way I see to do this is to write the expression out and set each of the terms being multiplied equal to zero. So we have: (x - 2)(2x - 5) = 0 (x - 2) = 0 (2x - 5) = 0 x = 2 2x = -5 x = -5/2 So when x = 2 the first term is zero and it makes the whole expression zero by multiplication. Likewise, when x = -5/2, the second term is zero and it makes the whole expression zero by multiplication. The reason that these are the only values that make this expression zero is because they are the only values that make at least one of the two terms zero. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Just as in the previous question, all three factors are being multiplied, this means that if an x value makes any one of these factors zero, the whole expression is, in turn, zero. So we have: (3x - 6) = 0 x + 4 = 0 (x^2 - 4) = 0 3x = 6 x = -4 x^2 = 4 X = 2 x = +-2 (when you take the square root you get to possible solutions here) So this expression (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = +2, -2, -4 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The trapezoid that is formed from the straight line from (3,5) to (7,9) is a sideways trapezoid when drawn. To find the area of the trapezoid I need the length of the two bases and the length of the height. Since it is sideways, I was able to derive the height by taking the difference between the two x-values. And the two bases are the two y-values of the points. So for the first trapezoid we have the area: h = 4 base#1 = 5 and base#2 = 9 Area = (1/2)(h)(base#1 + base#2) Area = (1/2)(4)(5 + 9) = (1/2)(4)(14) = (1/2)(56) = 56/2 = 28 units^2 The second trapezoid was sideways on the graph as well and I was able to derive the required variables in the same manner as above with the first trapezoid. The height was found taking the difference between the two x-values of the points and is 40 and the length of the two bases are again the two y-values. So for the second trapezoid area we have: h = 40 base#1 = 2 and base#2 = 4 Using the same area formula above I got… Area = (1/2)(40)(2 + 4) Area = (1/2)(40)(6) = (1/2)(240) = 120 units^2 So the second trapezoid has the larger area. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critic: ???I was able to attain the answer above, although I used different reasoning than the given solution provided and perhaps did too much for the problem; although I left it this way to ask if this is valid reasoning for this type of problem as well???
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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. ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As we move from left to right the graph increases as its slope increases is true the graph of y = x^2 on the interval x > 0. From the below table it can be seen that the function is increasing and its slope is getting larger and larger as it increases. x y 0 (0)^2 = 0 1 (1)^2 = 1 4 (4)^2 = 16 As we move from left to right the graph decreases as its slope increases is true for the graph of y = 1/x on the interval x > 0. From the table below it can be seen that the graph of this function is clearly decreasing although it decreases less and less we move to the right. Thus the slope is increasing in that it is getting close to 0 form the negative side. x y 0 1/0 (does not exist but approaches infinity) 1 1/1 = 1 4 1/4 = 0.25 As we move from left to right the graph increases as its slope decreases is true for the graph of y = sqrt(x) on the interval x > 0. From the table below, we can see that the function is increasing from left to right, although the increases are getting smaller and smaller. This means that the slope is decreasing to the right as the function increases to the right. x y 0 sqrt(0) = 0 1 sqrt(1) = 1 4 sqrt(4) = 2 9 sqrt(9) = 3 As we move from left to right the graph decreases as its slope decreases is not true for any of the above functions on the interval x > 0. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to think very hard on this one and put together a few things that I was not sure about. I’m not certain that I could answer a more in-depth problem such as this without review of this topic. ??? If there is anywhere you could suggest for me to get some extra practice on this topic that would be great??? ------------------------------------------------ Self-critique 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 . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The question asks for the energy of motion of an automobile at clock time t = 5. To do this I found the velocity at t = 5 V = 3(5) + 9 = 24 I then plug in velocity at this time to the energy of motion equation. E = 800 * 24^2 E = 800 * 576 E = 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. • ********************************************* Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes; E = 800(v)^2 V = 3t + 9 E = 800(3t + 9)^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All terms are being multiplied, this means that if any one of them is zero, they are all zero so we may solve each of the three expressions separately for zero. For the first expression: For the second expression: For the third expression 2^x - 1 = 0 x^2 - 25 = 0 2x + 6 = 0 2^x = 1 (x + 5)(x - 5) = 0 (factors) 2x = -6 (x)ln2 = ln1 x = -5 x = 5 x = -6/2 = -3 X = ln1/ln2 = 0 Solutions: x = 5, -5 Solution: x = -3 Solution: x = 0 So the above total expression is zero at x = 0, 5, -5, and -3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both of the trapezoids sketched here span from x = 3 to x = 7. This means they have the same width on the coordinate plane. However, The second trapezoid that spans from point (3,10) to (7,6) has a higher average altitude of 8 where as the first trapezoid only has an average altitude on the coordinate plane of 7. For this reason I am able to conclude that the second trapezoid that spans from point (3,10) to point (7,6) has the greatest area without necessarily calculating are as I did with a similar problem before. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Year one amount = $1000(0.10) + $1000 = $1100 Year two amount = $1100(0.10) + $1100 = $1210 Year three amount = $1210(0.10) + $1210 = $1331 Since the above pattern of equations equate to simply multiplying the original amount by 1.1 each year a more general expression can be derived. For the amount of money that is left after 40 years. Year forty amount = $1000 * 1.1^40 And for the general expression of t years Year t amount = $1000 * 1.1^t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!