#$&*
course Mth 173 8 / 27 11 : 20 Copy and paste this document into a text editor, insert your responses and submit using the Submit_Work_Form..............................................
Given Solution: OK `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: We know that anything multiplied to 0 is 0. Therefore for an expression in multiplicative form to be 0 either one or more factors should be 0. Given expression ( x - 2 ) * ( 2x + 5 ) For the expression to be 0 either of one or both of the factors should be zero. Thus either x - 2 or 2x + 5 should be 0 For x - 2 to be zero x should be 2 but for the same x value 2x + 5 is not zero For 2x + 5 to be zero x should be -5/2. For no other value of x either of the factor becomes 0 except for the values of x mentioned above and since there is no other factor the expression becomes 0 only at the 2 x values that are 2 and -5/2. Other way to say is when you combine the factors you get (x-2)(2x + 5) = 2x^2 + 5x - 4x - 10 = 2x^2 + x - 10 Since it is a binomial expression it has only two factors of x at which it becomes 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: Given expression ( 3x - 6 ) * ( x + 4 ) * ( x^2 - 4 ) which on expanding becomes (3x - 6 ) * ( x^3 + 4x^2 - 4x - 16 ) = 3x^4 + 6x^3 - 36x^2 - 24x + 96 Since it is a 4 degree equation it will have a maximum of 4 roots( values of x at which the expression becomes 0 ) As we know anything multiplied by 0 is 0. Thus for the entire expression to be 0 1 or more factors should be 0 So the expression is 0 when 3x - 6 is 0 3x - 6 = 0 x = 6/3 = 2 or x + 4 = 0 x = -4 or x^2 - 4 = 0 x^2 = 4 x = 2 or x = -2 so we obtain 4 values of x at which the expression is zero but out of the four there are 2 values which are same. Thus there are 3 values of x at which the expression is 0 which are 2 , -4 and -2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: Given first line passing through the points ( 3 , 5 ) and ( 7 , 9 ) Now when a line segment is drawn from these points to the x axis it meets the x axis as Line segment from ( 3 , 5 ) meets x-axis at ( 3 , 0 ) and line segment from ( 7 , 9 ) meets the x axis at ( 7 , 0 ) The obtained four co-ordinates of the first trapezium are ( 3 , 5 ) , ( 7 , 9 ) , ( 7 , 0 ) and ( 3 , 0 ) . Area of the trapezium calculated = (˝)*( sum of the length of the 2 parallel line)*( distance between the parallel lines) = (˝)*( 5 + 9 )*( 7 - 3 ) = (˝)*(14)*(4) = 28 square units Area of the first trapezium is 28 square units. The second line passing through ( 10 , 2 ) and ( 50 , 4 ) The line segment drawn from every points to the x axis meets the x axis as Lin segment drawn from ( 10 , 2 ) meets x axis at ( 10 , 0 ) and the line segment from ( 50 , 4) meets x axis at ( 50 , 0 ) The co-ordinates of the second trapezium obtained are ( 10 , 2 ) , ( 50 , 4 ) , ( 50 , 0 ) and ( 10 , 0 ) Area of the trapezium = (˝)*(sum of length of the 2 parallel line)*(distance between the parallel lines) = (˝)*( 4 + 2 )*( 50 - 10 ) = (˝)*(6)*(40) = 120 square units Area of the second trapezium is 120 square units. Thus the second trapezium has greater area..............................................
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: Given : The first graph y = x^2 The graph turns out to be a parabola whose lowest point is ( 0 , 0 ) . As we move from left to right for x > 0 the value of y increases as the slope also increases, The increase in the value of consecutive y increases for the same increase in x which also confirms the increase in slope. The second graph is y = 1/x For this graph the value of y decreases with increase in value of x greater than 0. The slope for the same graph is negative which is indicated by the decreasing value of y. The decrease in the value of y decreases as we move forward with the same increase in x and so the slope becomes more and more gradual and greater in value and lesser in magnitude approaching to 0. The third graph, which is y = sqrt(x) The graph again turns out to be a curve with increasing values of y with increasing values in x. The increase in the value of y decreases as we move forward with the value of x which indicates that the slope gradually decreases and approaches to a zero value. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: Initial population in the pond = 20 Percentage increase each month = 10% Increase in the number of frogs in the first month = 0.1* 20 = 2 Total frogs at the end of first month = 20 + 2 = 22 Increase in the number of frogs in the second month = 0.1 * 22 = 2.2 Total frog at the end of second month = 24.2 Increase in the number of frogs in the third month = 0.1 * 24.2 = 2.42 Total frog at the end of third month = 26.62 We can see that at the end of first month the total frogs = initial number * 1.1 At the end of second month = number of frogs at the end of first month * 1.1 = initial frogs * 1.1 * 1.1 Following the same procedure the total number of frogs at the end of 300 months would turn out to be = initial frogs * 1.1^300 = 20 * 1.1^300 = 2617010996188.39990701 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: 3 `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): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* 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: The value of 1/x for x = 1 , .1 , .01 , .001 turn out to be For x = 1 1/1 = 1 For x = .1 1/.1 = 10 For x = .01 1/.01 = 100 For x = .001 1/.001 = 1000 We can definitely see a pattern in the result obtained which is the number starts becoming larger and larger. The results gets multiplied by a factor of 10 to the previous result for a division of x by a factor of 10 to the previous x value. We say that the value of x is approaching to zero because the value of x continuously decreases its value by a factor of 10 from 1. As the magnitude of the x value decreases continuously it can be said that the value of x approaches 0 For x to continue approaching 0 we must continue dividing it by a factor of 10 or more, which turns out to go on as .0001 , .00001 , .000001 , .0000001 . as the number of zeros after the decimal and before the 1 increase the value decreases and approaches 0. The values of 1/x increases with decrease in value of x as stated earlier. Now as the value of x approaches zero the value of x starts becoming larger and larger and start approaching infinity. The graph of 1/x vs x for values of x from 0 to 1 turns out to be a graph with 1/x approaching to infinity with x approaching to 0 and 1/x approaching to 0 with x approaching to infinity. The graph continuously decreases with its slope always negative and approaching to zero. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: Given: Equation for velocity and time : v = 3t + 9 Equation for energy and velocity : E = 800v^2 Time at which energy is to be calculated : t = 5 Since there is no direct relationship between energy and time therefore from the time we need to calculate the velocity and then the energy. Velocity at t = 5 : v = 3*5 + 9 = 15 + 9 = 24 And thus the energy at the obtained velocity that is at time 5 :E = 800*24*24 E = 460800 units Ans : Thus the total energy of the automobile at clock time t = 5 is 460800units confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800. • ********************************************* Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Expression for E in terms of t can be obtained by substituting v in the energy time equation for v obtained in velocity time equation. Given equations v = 3t + 9……(1) and E = 800v^2…….(2) Substituting (1) in (2) we get E = 800*(3t + 9)^2 E = 800*(3t + 9)*(3t + 9) E = 800*(9t^2 + 54t + 81) E = 7200t^2 + 43200t + 64800 Ans : E = 7200t^2 + 43200t + 64800 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: Again the expression becomes zero at its roots, that is when 1 or more factors of the multiplicative part becomes zero. Given expression: ( 2^x - 1 ) * ( x^2 - 25 ) * ( 2x + 6 ) On expanding ( 2^x - 1 )*( 2 x^3 + 6x^2 - 50x - 150 )………..(1) The equation on the right part of the expression…….(1) is cubic thus at the most there must 3 values of x as roots and a factor on the left part of the expression……..(1) add up to a total of at most 4 values of x. Now the left part of the expression…...(1) 2^x - 1 = 0 2^x = 1 x = 0. Now for the next factor to be zero which is x^2 - 25 x^2 = 25 x = 5 or x = -5 For the 3rd factor which is 2x + 6 2x + 6 = 0 x = -3 so finally the values of x at which either 1 or more factors becomes zero making the expression zero are 0 , 5 , -5 , -3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: One straight line segment connects the points (3,5) and (7,9) while another connects the points (3, 10) and (7, 6). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Any solution is good, but a solution that follows from a good argument that doesn't actually calculate the areas of the two trapezoids is better. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the 1st trapezium the co-ordinates given are ( 3 , 5 ) and ( 7 , 9 ) and for the 2nd trapezium the co-ordinates given are ( 3 , 10 ) and ( 7 , 6 ) The run for both the trapezium is the same thus the area for the majority is not affected by that factor. The first co-ordinate of the second trapezium that is ( 3 , 10 ) is at a greater height as compared to the co-ordinate ( 3 , 5 ) of the first trapezium by a factor of 5 units. The second co-ordinate of the first trapezium that is ( 7 , 9 ) is at a greater height as compared to the co-ordinate ( 7 , 6 ) of the second trapezium by a factor of 3 units. The difference in the height values definitely indicate the reason for the majority in areas and since 5 units difference is greater than 3 units difference in height , thus the first factor decides the majority. Since ( 3 , 10 ) is at a greater height to ( 3 , 5 ) thus the trapezium with ( 3 , 10 ) and ( 7 , 6 ) that is the 2nd trapezium has a greater area. confidence rating #$&*: 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: Initial investment = $1000 Percentage of amount added every year = 10% of the total amount the previous year Amount added at the 1st year = .1*1000 = $100 Total amount at the end of 1st year = $1100 Amount added at the 2nd year = .1*1100 = $110 Total amount at the end of 2nd year = $1210 Amount added at the 3rd year = .1*1210 = $121 Total amount at the end of 3rd year = $1331 We can see that the total amount at the end of any year = 1.1*the total amount of the previous year = 1.1*1.1* the total amount of the year previous to the previous year Thus continuing the same trend the total amount at the end of 40 years =1000*((1.1)^40) = 45259.2555 Similarly the total amount at the end of t years = 1000*((1.1)^t) And thus the total amount earned at the end of t years = total amount at the end of t years - initial investment = 1000*((1.1)^t) - 1000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!