#$&* course Mth 164 1/15 4:24 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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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: The reason why (x-2) * (2x+5) is zero when X= 2 or -2.5 is because when you set x - 2 = 0 or 2x + 5 = 0 and solve for X you get 2 or -2.5. Anything times by 0 is going to be 0. X - 2 = 0 add 2 to both sides X = 2 This means 2 is a root or zero of the graph y = x - 2 2x + 5 = 0 Add -5 to both sides 2x = -5 Now divide by 2 x = -5/2 or 2.5 once again this is a zero of the graph y = 2x + 5 Now entering the X’s into the formula (x-2) * (2x+5). X=2 first (x-2) * (2x+5) Enter 2 in for x (2-2) * (2*2+5) do what is in grouping symbols 0 * 9 multiply (anything times 0 equals 0) 0 Now for x = -2.5 (x-2) * (2x+5) Enter 2.5 for x (2.5-2) * (2*-2.5+5) Do what is in grouping symbols .5 * 0 multiply (anything times 0 equals 0) 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 0I 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 - acbut 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find what x values will be zero you need to set each part equal to 0. 3x - 6 = 0 add 6 to both sides 3x = 6 divide by 3 on both sides x = 2 is one of the x’s that will make (3x - 6) * (x + 4) * (x^2 - 4) be zero. x + 4 =0 add a -4 to both sides x = -4 is another x that will make (3x - 6) * (x + 4) * (x^2 - 4) be zero. x^2 - 4 = 0 add 4 to both sides x^2 = 4 now take the square root of both sides x = -2, 2 any time you take a square root there is a possibility of two answers. In this case -2 and 2 are both answers. Our final x’s are x= -4, -2, 2 x =-4 (3x - 6) * (x + 4) * (x^2 - 4) Enter -4 in for x (3*-4 - 6) * (-4 + 4) * (-4^2 - 4) do what is in grouping first -18 * 0 * 12 multiply the numbers (anything times 0 equals 0) 0 x = -2 (3x - 6) * (x + 4) * (x^2 - 4) Enter -2 in for x (3 * -2 - 6) * (-2 + 4) * (-2^2 - 4) do what is in grouping -12 * -2 * 0 multiply the numbers (anything times 0 equals 0) 0 x = 2 (3x - 6) * (x + 4) * (x^2 - 4) Enter 2 in for x (3 * 2 - 6) * (2 + 4) * (2^2 - 4) do what is in grouping 0 * 6 * 0 multiply the numbers (anything times 0 equals 0) 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): ------------------------------------------------ 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: To see which which trapezoid is bigger you can use the distance formula d = sqrt( (x2-x1)^2 - (y2-y1)^2) Points (3,5) and (7,9) d = sqrt( (x2-x1)^2 - (y2-y1)^2) Enter the x’s and y’s d = sqrt( (7-3)^2 - (9-5)^2) do what is in grouping d = sqrt( (4)^2 - (4)^2) do what is in grouping d = sqrt( 16- 16) do what is in grouping d = sqrt( 0 ) take the square root 0 d = 0 Points (10,2) and (50, 4) d = sqrt( (x2-x1)^2 - (y2-y1)^2) Enter the x’s and y’s d = sqrt( (50 - 10)^2 - (4 -2 )^2) do what is in grouping d = sqrt( (40)^2 - (2)^2) do what is in grouping d = sqrt( 1600 - 4) do what is in grouping d = sqrt( 1596 ) take the square root 1596 d = 39.949 This proves that the second trapezoid has a bigger area than the first one because the distance between the points in the second one is bigger than the first one. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. The graph y = x^2 best fits. As we move from left to right the graph increases as its slope decreases. The graph y = sqrt(x) best fits As we move from left to right the graph decreases as its slope decreases. The graph y = 1 /x best fits. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: After the first month you would have 22 frogs because 20 * 10% = 2 and 20 + 2 = 22. After the second month you would have 24 frogs because 22 * 10% = 2.2 but you cannot have .2 of a frog so it is just 2. After the third month you would have 26 frogs because 24 * 10% = 2.4 but once again you cannot have .4 of a frog. 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): To calculate who to see how many frogs you would have after 300 months I should have used powers. 20 *10%^300 ------------------------------------------------ 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: X y 1 1 .1 10 .01 100 .001 10000 The pattern is as x gets smaller and smaller the y grows larger. The reason we say that the x is approaches 0 is because as it the number get small they are closing to be 0. Some more numbers we could use are .00001, .000001 and so on. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 5 is a given to use v = 3 t + 9 Enter 5 for t v = 3 *5 + 9 v = 15 + 9 v = 24 so now we will put our v into the E = 800 v^2 formula v = 24 E = 800 v^2 Enter 24 for v E = 800 * 24^2 do 24^2 E = 800 * 576 multiply E = 460800 confidence rating #$&*: 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: E = 800 ( 3t + 9) ^2 This reason for this is be cause the 3t + 9 is what gave us our v to begin with so why not go ahead substitute it in for v. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 - 25 = 0 x^2 = 25 take square root of both sides x = -5 and 5 2x + 6 = 0add -6 to both sides 2x = -6 divide by 2 x = -3 2^x - 1 = 0 add 1 to both sides 2^x = 1 rewrite 2^x as log x = 2i /log 2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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: The area of both of the trapezoids are the same because the distance between the points are both the same. Points (3,5) (7,9) 7-3 = 4 9 - 5 = 4 points (3, 10) (7,6) 7 - 3 = 4 10 - 6 = 4 both are the same.