#$&* course mth163 1/21 - 9:04pm Copy and paste this document into a text editor, insert your responses and submit using the Submit_Work_Form.
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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: ********************************************* 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: Verification of claims for x=2 and x= -2.5: (2-2) * (2*2 + 5) 0 * (4+5) NOTE: Given the question it is not neccessary to go further with this problem, although I will finish the solution. We have here zero times the operations in parenthesis. zero times anything is zero. 0 * 9 0 (-2.5-2) * (2*-2.5 + 5) -4.5 * (-5 + 5) -4.5 * 0 0 This first equation can only be brought to 0 when one side (either set of parenthesis in this case) of the equation calculates to zero before being multiplied by the other side. It is ONLY necessary for ONE side to become zero. In this case, the only way to do so is by subtracting 2 from 2 to create zero. In the second equation, we must come up with -5 to balance out the 5 in our effort to create zero. The only way to create -5 on that side of the equation is to multiply 2 by -2.5, which gives us 5. 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 ñ ac but I donít 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: ********************************************* Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Again, one of the three units in this equation must be made equal to zero in order to produce a final solution of zero. The x values that do so are: 2, and -4 shown as follows x=2 applies to both (3x-6) and (x^2-4): (3*2 - 6) = (6-6) = 0 (2^2 - 4) = (4 - 4) = 0 x= -4 applies to (x+4) (-4 + 4) = 0 ADDITION AFTER READING GIVEN SOLUTION. EXPLANATIONS FOUND IN CRITIQUE SECTION. 3x - 6 = 0 3x = 6 x = 2 2 is a solution that produces zero. x+4 = 0 x = -4 -4 is a solution that produces zero. x^2 -4 = 0 x^2 = 4 x= +/- the SQRT of 4. 2 is already a solution we found above, therefore we can add -2 to the list. SELF REMINDER: The answer is + or - Because the square root of something is asking: “what can be multiplied together twice in order to produce this number?” Remember that two negatives multiplied result in a positive. Thus why -2 is part of this answer. 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 initially completed this problem without setting each equation equal to zero, as the math seems simple at face value. However this got me into trouble with the squared value, as I didn’t initially remember the + or - that could come from solving that equation properly. ------------------------------------------------ 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: INITIAL LOGIC: Count units to determine a rough area of each trapezoid, as the math is difficult at this point. There are roughly 19 1x1 units in the first (vertical) trapezoid. There are 40 units in the bottom level of the second (long horizontal) trapezoid alone. As the top slants, it is more difficult to determine the number of units. There are certainly more than double the units, which makes the second one the larger of the two trapezoids. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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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: point 1 is correct for the graph of y = x^2. as it increases, its slope increases as well because the value of y goes up a larger amount with each increase in x. Point 2 is correct for the graph of y = 1/x. as it decreases, the value of y gets closer and closer to 0, but will never quite get there. (1/1, 1/2, 1/3, 1/4, 1/5 etc…) Therefore the slope gets less negative, thus, increasing Point 3 is correct for the graph of y = SQRT(x). as it increases, the value of y goes up by a smaller increment over time. Thus, the slope decreases as x rises. Point 4 is not correct for any graph. 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 0: 20 frogs Month 1: (20 * .1) + 20 = 2 + 20 = 22 Frogs (we add 20 again because (20*.1) will give us what 10% of 20 is. we then have to add that to the base population of 20 Month 2: (22 *.1) + 22 = 2.2 + 22 = 24.2 Frogs Month 3: (24.2 *.1) + 24.2 = 2.42 + 24.2 = 26.62 Frogs. There are 26.62 frogs at the end of the third month. My initial reaction is to set up an equation as such: 20 * .1^x, where x is the number of months you want to determine. However this leaves out the important aspect of adding back in the original value, and I am stuck. UPDATE: I was at least in the correct mindset, but I did not think about using 1.1 instead. I will put my critique in the appropriate section below. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1.5
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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): The only issue I had with this problem was setting up an equation to determine the number of frogs after 300 months. As mentioned above, I wrote down 20 * .1^x, where x is the number of months you want to determine, but that is as far as I could wrap my head around it. Using 1.1 makes sense because it INCLUDES the base value of frogs in the equation itself (the 1 in the ones value spot). In my own words: When multiplying by a fraction less than one, or a decimal, if you want to INCLUDE the base value in your solution (have it added back in), then add 1 in the ones spot, whether it be in decimal form, or fraction form. This essentially just multiplies the original number (in this case, frogs) by itself, and then multiplies and adds to it the decimal point or fraction (in this case, .1 for 10%). NOTE: This logic would not apply to something along the lines of “doubling”. I.E. You would not multiply by 2.1 if you wanted to know what double the amount of frogs in the beginning would have produced. You would have to do the math to determine 40*(1.1^300). THIS PREVIOUS NOTE APPEARS HERE MAINLY FOR MY OWN CLARIFICATION AND FUTURE REFERENCE. ------------------------------------------------ Self-critique Rating: ********************************************* 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: 1/x for the requested values: 1/1 = 1 1/.1 = 10 1/.01 = 100 1/.001 = 1000 The given values for x were: 1, .1, .01, and .001. Each of these approaches closer and closer to zero because each number is one-tenth smaller than the preceding number. However, this pattern can continue indefinitely, without ever actually reaching zero. The next numbers in this series that we would use to continue approaching zero would be: .0001, .00001, .000001, etc.. The values of the solution to 1/x as x gets closer to zero increase. As you can see above, each time the number increases by 10 times with each step. It is interesting to note that with each drop in the value of x by 1/10th of itself, the value of the solution to 1/x increases by 10-times itself. ????? I will describe the graph in two ways below, the first being how it would look to someone second-hand, and the second taking into account the ORDER in which the points were plotted. My assumption is that the former is the proper way in which to describe this graph. But I am not completely positive as to whether or not you must describe a graph in the order in which they were plotted. 1) In terms of face value, this graph is decreasing, very rapidly. The slope of the graph increases. “This graph decreases while the slope increases”. Again, so that it can be corrected should I be wrong, to put in simple terms, as the graph decreases, the slope BECOMES LESS NEGATIVE, thus, increases. As you look up the y axis of the graph, the values of x would get very close to 0, but would never actually touch the axis.
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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. However please note the question above that is within with my solution. I understand the graph from two perspectives (looks, and order of plotted points), and am unsure which way is the correct way to interpret this graph. If this involved motion of a physical object, then I would assume it greatly matters! ------------------------------------------------ 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: In order to determine energy, we must first determine velocity. v=3t+9. For which we will substitute a clock time (elapsed time) of 5. v=3*5 +9 v=15+9 v= 24 Now that we have determined the velocity (24), we can use that to determine the energy. E = 800v^2 E = 800*24^2 E = 800*576 E = 460,800 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800. ï ********************************************* Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We need to adapt the E = 800v^2 to contain t instead of v. Simply put, we are just going to insert the equation for v in place of the letter v. It yields / means the same thing, it is just not expressed in terms of “velocity”, it is instead expressed as a full equation. E = 800(3t + 9)^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: ********************************************* Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero? We need to have one of each set equal to zero in order to produce a full-equation solution of zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2^x - 1 = 0 :In this equation there is only one option. we need for 2^x to be equal to 1, so that we end up with 1-1 = 0. The ONLY way to do this is to have x=0, as anything to the power of zero = 1. x^2 - 25 = 0 x^2 = 25 x = +/- 5 2x+6 = 0 2x = -6 x = -3. The values for x that produce a solution of zero for this expression are: 0, +/-5, -3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ********************************************* Question: One straight line segment connects the points (3,5) and (7,9) while another connects the points (3, 10) and (7, 6). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Any solution is good, but a solution that follows from a good argument that doesn't actually calculate the areas of the two trapezoids is better. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each of these triangles forms a trapezoid with sides lining up on X= 3, and X = 7, Both therefore have the same width. If we think of these trapezoids as squares with triangles placed on their tops, then we can pretty easily figure out which has the greater area: Trapezoid 1 (3,5) (7,9) has a SQUARE (referring to shape, not operation) width of 4, and a height of 5. The triangle on top of the square then extends up 4 units to a height of 9. Trapezoid 2 has the same square width and height. It’s triangle however extends up 5 units to a height of 10. Because both trapezoids have the same “square shape” area, the one with the higher / larger triangle is the one with the greater area. Trapezoid 2 has the greater area. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ********************************************* Question: Suppose you invest $1000 and, at the end of any given year, 10% is added to the amount. How much would you have after 1, 2 and 3 years? What is an expression for the amount you would have after 40 years (give an expression that could easily be evaluated using a calculator, but don't bother to actually evaluate it)? What is an expression for the amount you would have after t years? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Year 0: 1000 Year 1: (1000 * .1) + 1000 = 1100 (we must add the 1000 base back in after we determine what 10% of it is in the first half of the equation. Year 2: (1100 * .1) + 1100 = 1210 Year 3: (1210 * .1) + 1210 = 1331 From our previous question, we learned that we can set up an equation to determine the amount we would have after any number of years. Without repeating the logic behind it, as I explained earlier, the equation would be: 1000 * 1.1^40 If we don’t care about a specific number of years, we can just replace the number 40 in the equation above with the letter t: 1000 * 1.1^t where t = years. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!