#$&* course Mth 158 1/14 1:30pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating:3 ********************************************* Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I evaluate the expression using x = 2: ((2)-2) * (2(2)+5) = 0 * (4 + 5) = 0 * (9) = 0*9 = 0 Because any number multiplied by 0 equals 0, x=2 will solve for 0. By inputting 2 for x in the first parenthetical set and evaluating, we end up with 0. All other numbers including -2.5, entered into this first set leave us with a number other than 0. Evaluating the expression using x= - 2.5: ((-2.5)-2) * (2(-2.5)+5) =(-4.5) * (-5 +5) =-4.5 * 0 =0 Again any number multiplied by 0 equals 0, so x= -2.5 is evaluated out to 0. By putting -2.5 instead of x into the expression, the second parenthetical set evaluates out to 0. All other numbers, including 2, when entered into this set equal a number other than 0. Therefore only 2 and -2.5 will leave one or the other parenthetical sets equaling 0. confidence rating #$&*: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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand the distributive law and how it affects this expression. Self-critique Rating:3 ********************************************* Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the x-values that will make the expression be zero, one of the parenthesized sets must equal zero, since multiplying the other sets by zero will in turn equal zero. (3x -6)=0, (x +4)=0 or (x^2 -4)=0. (3x – 6) = 0 3x = 6 x = 6/3 x =2 (x + 4) = 0 x=-4 x^2 – 4 = 0 x^2 = 4 'sqrt x^2 = 'sqrt 4 x = +2,-2 If x = 2, (3(2) – 6) * ((2)+4) * ((2)^2 – 4) = (6 – 6) * (6) * (4 -4) = (0) * (6) * 0 = 0. This checks out because the product of the expression when x = 2 is zero. If x = -4,(3(-4) – 6) * ((-4)+4) * ((-4)^2 – 4) =(-12 – 6) * (0) * (16 - 4) =(-18) * (0) * (12) =0 Again, this solution also checks out because the product of the expression when x = -4 is zero. When x = -2, (3(-2) -6) * ((-2) + 4) * ((-2)^2 - 4) = (-6 – 6) * (2) * (4 - 4) = -12 * 2 * 0 = 0 Again we are left with 0 so -2 is also a solution to the expression. 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): I found the same solution to be true. Self-critique Rating: ********************************************* 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 created by the points (10,2) and (50, 4) has an area of 120 units sq. The trapezoid created by the points (3,5) and (7,9) has an area of 28units sq. The trapezoid created by the points (10,2) and (50, 4) therefore has a larger surface area. I found this by completing the trapezoids as instructed and drawing a line along the x-axis at the lower of the two points in each line segment. Eg: from (10,2) to (50,2) for the second line segment. This way I was able to calculate the area of a rectangle 2 units high by 40 units wide = 80 units sq. I then drew a line from (50, 4) to (10, 40) to create another rectangle with 80 units sq. Dividing this in half gives me the area of the top portion of my trapezoid ( points (10, 2), (50, 2) and (50, 4). By adding the area of my first rectangle with the area of this triangle, I arrived at the area of the trapezoid which = 120 (80 + 40 = 120 units sq.) Using the same strategy, I found the area of the first trapezoid to be 28 units sq. 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. ********************************************* 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: For x^2, the graph decreases at a decreasing rate until x=0, after x=0 the graph starts to increase at an increasing rate. For 1/x, The graph decreases at a decreasing rate, the greater the x value is. For 'sqrt(x), the graph increases at a decreasing rate the greater the value of x is. 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: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Month 1 : (.1 * 20) + 20 = F 2 + 20 = F 22 = F Month 2: (.1 * 22) + 22 = F 2.2 + 22 = F 24.2 = F Month 3: (.1 * 24.2) + 24.2 = F 2.42 + 24.2 = F 26.62 = F 26.62 – 24.2 = 2.42 24.2 – 22 = 2.2 22 – 20 = 2 The growth in the first month is 2 frogs. 2 * .1 (rate of increase) = .2 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was OK with the problem until I tried to find a formula to figure frog growth for any month in this case 300. After seeing the given solution it made sense to me, I was trying to figure out that rate, but was unsure how to come up with the constant which I now see is 1.1. Instead of multiplying .1 by the difference in each month, I should have divided one months frogs by the previous months frogs (26.62 / 24.2 = 1., 24.2 / 22 =1.1, etc.) Self-critique Rating:3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Calculating 1/x for the values given shows a pattern where the closer x gets to zero, the greater the value of y. This is because when solving 1/.01 we are looking at a fraction of 1/(1/10) to solve this, we need to multiply by the inverse, or 1/1 * 10/1 = 10. Since 1/1 = 1, 1/.10 is going to prove the larger number. The graph to this expression would show a graph sharply decreasing in y values, until x=1, then it would taper off to an almost flat line (2,.5), (3,.33), (4,.25),(10,.1) We would say the graph is decreasing at a decreasing rate. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I was saying the same thing, however in the opposite direction, you described the line as it approached the y axis, I was describing the line as it approached the x-axis. In both cases, the line will never touch the y or x-axes. Self-critique Rating:3 ********************************************* Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t=5: v = 3(5) + 9 = 15 + 9 = 24 E = 800 * (24)^2 = 800 * 576 = 460800 Therefore E of the automobile is 460800 when v = 24 and t =5. confidence rating #$&*: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: 800 * (3t + 9)^2 Check: 800 * (3(5) + 9)^2 = 800 * (15 +9)^2 = 800 * (24)^2 = 800 * 576 = 460800 =check 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique 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 #(*!