#$&* course Phy 202 1/18/158:55 AM 003. PC1 questions
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Given Solution: `aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,19) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, let’s substitute the values of x=2 and x=2.5 into the equation. For x=2: (x-2) * (2x+5) (2-2) * (2(2)+5) (0) * (4+5) = 0 (0) * (9) = 0 Because we are multiplying and the resulting value for the first set of parentheses (x-2) was 0, when we multiply it to the other set of parentheses, we get an overall value of 0. For x=-2.5 (x-2) * (2x+5) ((-2.5)-2) * (2(-2.5)+5) (-4.5) * ((-5)+5) (-4.5) * (0) = 0 For this equation, using the value x=-2.5 results in the parentheses (2x+5) becoming 0, and since we are again multiplying, our overall resulting value is 0. Only the values x=2 and x=-2.5 can result in the parentheses becoming equal to 0 and provide and overall value of 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ok
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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: To begin this problem, and determine what values x must be in order to get an overall value of 0, we can look at the different set of parentheses that are involved with the equation individually. Since we know that we are multiplying, as long as we have a value of 0 for one the parentheses, our overall equation will equal 0. Perhaps one of the easiest ways to do this is to set each set of parentheses equal to 0 and see what values we get. For (3x - 6): (3x-6) = 0 3x = 6 X = 6/3 = 2 For (x + 4): (x + 4) = 0 X = -4 For (x^2 - 4): (x^2 - 4)=0 (x^2) = 4 Square root [(x^2)] = square root [4] X=2 In order to get an overall value of 0 for this equation, x=2 or x= -4. This is because these are the only two values that will result in the parentheses equaling 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not catch that x = -2 for the final parentheses, (x^2 - 4), would also result in the value of 0. However, it is very obvious when looking at the equation. I had simply overlooked this part of it. I now realize that anytime we are taking a square root of a value, we can have either the + or - value of it. This is because when we raise a value to any even number power, it does not matter whether or not the value is positive or negative…we will always get a positive value. (x^2 - 4) ((-2)^2 - 4) (4-4) = 0 ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: We know that Area = length * width. So all we must do for these two trapezoids is determine their length and width of find which one will have the greater area. For Trapezoid (3,5) and (7,9): Width = difference in x values Width = 7-3 = 4 units Length = Average of y values Length = (9+5)/2 = 7 units Area = Length * Width Area = (7units*4units) = 28 units^2 For Trapezoid (10,2) and (50,4): Width = difference in x values Width = 50 - 10 = 40 units Length = Average of y values Length = (4+2)/2 = 3 Area = Length * Width Area = (3 units*40 units) = 120 units^2 The greater area belongs to the trapezoid formed by the points (10,2) and (50,4). 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: As we move from left to right the graph increases as its slope increases. Y = x^2 (As x values increase, the y values increase by a value x^2). This graph increases at an increasing rate. As we move from left to right the graph decreases as its slope increases. None of the graphs of the three equations match this description. As we move from left to right the graph increases as its slope decreases. y = `sqrt(x) (As the x values continue to increase, the y values increase as well, but with a much smaller difference). This graph is increasing at a decreasing rate. As we move from left to right the graph decreases as its slope decreases. y = 1/x (As the x value is increasing, the y value is decreasing. The y value will never equal 0, but will continue to get only closer and closer to the x axis.). This graph is decreasing at 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* 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: For month 1: 20 frogs (.10) = 2 frogs 20 frogs + 2 frogs = 22 frogs For month 2: 22 Frogs (.10) = 2.2 frogs 22 Frogs + 2.2 Frogs = 24.2 frogs For month 3: 24.2 Frogs (.10) = 2.42 frogs 24.2 frogs + 2.42 frogs = 26.62 frogs I am having difficulty coming up with an equation to determine how many frogs will be present in the pond after 300 months. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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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): After reading the given solution, I can easily see how multiplying by 1.1 is the exact same as multiplying a number by .1 and adding it to the number. I can also see how by raising 1.1 to the power equal to the number of months and multiplying that to the beginning 20 frogs provides you with an answer to how many frogs will be present within that number of months. For example: For four months: 20 frogs x (1.1^4) 20 frogs x (1.4641) 29.282 frogs For 300 months: 20 frogs x (1.1^300) 20 frogs x (2.617011*10^12) 5.234022*10^13 frogs ------------------------------------------------ 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: For equation 1/x: (1/1) = 1 (1/.1) = 10 (1/.01) = 100 (1/.001) = 1000 As the value of x decreases (becomes closer to 0) the overall value of the equation continues to increase. We say that the values of x are approaching zero because they are becoming less and less, and are decreasing in numerical value. The way that we are decreasing x’s value in this problem is by dividing the previous x value by 10. Other x values we can use to have it continue to become closer to zero are x= .0001, .00001, and .000001. As the values of x approach 0, the values of 1/x continue to increase. For graphing this equation, the x values would continue decrease (become closer and closer to zero), and the y values would continue to increase. The graph would never have an x or y intercept, and would always have positive x and y values. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, let’s find v for t = 5: v = 3 t + 9 v = 3 (5) + 9 v = (15) + 9 v = 24 Now, we can substitute the value for v into the equation E = 800* v^2: E = 800* v^2 E = 800* (24)^2 E = 800 * 576 E = 460800 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: Since we know that (E = 800 v^2) and that (v = 3 t + 9), we can simply substitute (v = 3 t + 9) into (E = 800 v^2): E = 800 v^2 E = 800* (3 t + 9)^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To begin this problem, and determine what values x must be in order to get an overall value of 0, we can look at the different set of parentheses that are involved with the equation individually. Since we know that we are multiplying, as long as we have a value of 0 for one the parentheses, our overall equation will equal 0. Perhaps one of the easiest ways to do this is to set each set of parentheses equal to 0 and see what values we get. Since x is the exponent in the first set of parentheses, we know that the resulting value of 2^x must be equal to 1 so that we get (1-1) = 0. Because any number to the 0 power = 1, we know that x must equal 0. (2^x - 1)=0 (2^(0) - 1)=0 ((1) - 1)=0 X = 0 ( x^2 - 25 )=0 X^2 = 25 Sqrt(x^2) = + - sqrt(25) X = + - 5 (2x + 6) =0 2x = -6 X = -6/2 = -3 The solution for this equation will be 0 when x = 0,5, -5, and -3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: From looking at the points that are given for the two different trapezoids, we can tell that the trapezoid with the points (3, 10) and (7, 6) will have the greater area. This is because it has greater numbers for the y values than the first trapezoid. Both trapezoids have the exact same x values so we know that they will have the same width in this aspect. However, when looking at the y values, we can see that the second trapezoid has a greater height (length). Because of this we can tell that the trapezoid with the points (3, 10) and (7, 6) will have a greater area than the trapezoid with points (3,5) and (7,9). *I went ahead and performed the calculations to further support and show my reasoning* We know that Area = length * width. So all we must do for these two trapezoids is determine their length and width of find which one will have the greater area. For Trapezoid (3,5) and (7,9): Width = difference in x values Width = 7-3 = 4 units Length = Average of y values Length = (9+5)/2 = 7 units Area = Length * Width Area = (7units*4units) = 28 units^2 For Trapezoid (3,10) and (7,6): Width = difference in x values Width = 7-3 = 4 units Length = Average of y values Length = (10+6)/2 = 8 units Area = Length * Width Area = (8units*4units) = 32 units^2 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 1: $1000 (.1) = $100 $1000 + $100 = $1100 Year 2: $1100 (.1) = $110 $1100 + $110 = $1210 Year 3: $1210 (.1) = $121 $1210 + $121 = $1331 An expression for determining the amount of money after 40 years would be: $1000 * (1.1)^40 An expression for determining the amount of money you would have after t years would be: $1000 * (1.1)^t (t=#of years) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!