course mth158 9/10/09 @ 10:55pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.vvvv
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Given Solution: * * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | |4x| - |5y| | =| |4(3)| - |5(-2)| | =|12-10 | = 2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. ** * R.2.64 (was R.2.54) Explain what values, if any, cannot be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = (-9 (0^2) – 0+1) / (0^3 +0) = (-9-1) / 0 =-10/0 = undefined Any number divided by 0 is undefined confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 is, and only if, either x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2.76 \ 73 (was R.4.6). What is -4^-2 and how did you use the laws of exponents to get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = -4^-2 = -1 4^2 = -1/16 The exponent cant be negative so in order to make it a positive exponent you have to put 1/-4^2 which would make it -1/16. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** order of operations implies exponentiation before multiplication; the - in front of the 4 is not part of the 4 but is an implicit multiplication by -1. Thus only 4 is raised to the -2 power. Starting with the expression -4^(-2): Since a^-b = 1 / (a^b), we have 4^-2 = 1 / (4)^2 = 1 / 16. The - in front then gives us -4^(-2) = - ( 1/ 16) = -1/16. If the intent was to take -4 to the -2 power the expression would have been written (-4)^(-2).** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the answer correct however I did not go about getting that answer the same way as in the solution. I now realize the principles behind the solution. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: =3^-2 / 3^2 * 5^3 /5 = 3^(-2-2) *5^(3-1) =3^-4 * 5^2 =1/81 * 25 =25/81 The law I used was a^m / a^n= a^(m-n) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with (3^(-2)*5^3)/(3^2*5): Grouping factors with like bases we have 3^(-2)/3^2 * 5^3 / 5. Using the fact that a^b / a^c = a^(b-c) we get 3^(-2 -2) * 5^(3-1), which gives us 3^-4 * 5^2. Using a^(-b) = 1 / a^b we get (1/3^4) * 5^2. Simplifying we have (1/81) * 25 = 25/81. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2.94. Express [ 5 x^-2 / (6 y^-2) ] ^ -3 with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: =(5x^-2)^-3 (6y^-2)^-3 =5^-3(x^-2)^-3 6^-3(y^-2)^-3 =5^-3 x^6 6^-3 y^6 =-125x^6 -216y^6 =216x^6 125y^6 The theorem I used is (a/b)^n= (a^n) / (bn). I also referred to ex. 10 in section R2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: [ 5 x^-2 / (6 y^-2) ] ^ -3 = (5 x^-2)^-3 / (6 y^-2)^-3, since (a/b)^c = a^c / b^c. This simplifies to 5^-3 (x^-2)^-3 / [ 6^-3 (y^-2)^-3 ] since (ab)^c = a^c b^c. Then since (a^b)^c = a^(bc) we have 5^-3 x^6 / [ 6^-3 y^6 ] . We rearrange this to get the result 6^3 x^6 / (5^3 y^6), since a^-b = 1 / a^b. STUDENT QUESTION: I do not see how you can take and seperate the problem down like this has it seems to just have reversed the problem around in a different ordering and I do not see how this changed the exponets from being negative Is there anyway you can explain this problem in a little more depth INSTRUCTOR RESPONSE: A fundamental law of exponents is that exponentiation distributes over multiplication, so that (a * b) ^ c = a^c * b^c and (a / b) ^ c = a^c / b^c More specifically, if c = -3 then we have ( a * b ) ^ (-3) = a * (-3) * b^(-3) and ( a / b ) ^ (-3) = a ^ (-3) / b^(-3). Now a ^ (3) / b^(3) = 1 / a ^ (3) / (1 / b^(3)) and 1 / a ^ (3) / (1 / b^(3)) = 1 / a^3 * (b^3 / 1) = b^3 / a^3. This principle applies to any string of multiplcations and division, so for example ( a * b / (c * d) ) ^ e = a^e * b^e / (c^e * d^e). If e = -3 then we would have ( a * b / (c * d) ) ^ (-3) = a^(-3) * b^(-3) / (c^(-3) * d^(-3)). Since the -3 power is the reciprocal of the 3 power this expression becomes 1/a^(3) * (1/b^(3)) / (1/c^(3) * (1/d^(3))), which is easily seen to be equal to 1 / (a^3 * b^3) / (1 / (c^3 * d^3) ). Dividing by (1 / (c^3 * d^3) ) is the same as multiplying by (c^3 * d^3) / 1 so 1 / (a^3 * b^3) / (1 / (c^3 * d^3) ) = 1 / (a^3 * b^3) * (c^3 * d^3) = (c^3 * d^3) / (a^3 * b^3). You should have written the above expressions, which are difficult to read in this notation, on paper, applying the order of operations. The expressions you wrote down should look like the ones below. Be sure you understand the translation from the 'typewriter notation' above to the standard notation depicted below, and be sure you know how to write each of the expressions depicted below in standard notation: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = (-8)^-2 x^(3*-2) = 64 x^-6 =1/64 x^6 Law of exponents used: (a^m)^n=a^mn Exponents cannot be left negative: a^-n=1/a^n confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** ERRONEOUS STUDENT SOLUTION: (-8x^3)^-2 -1/(-8^2 * x^3+2) 1/64x^5 INSTRUCTOR COMMENT: 1/64x^5 means 1 / 64 * x^5 = x^5 / 64. This is not what you meant but it is the only correct interpretation of what you wrote. Also it's not x^3 * x^2, which would be x^5, but (x^3)^2. There are several ways to get the solution. Two ways are shown below. They make more sense if you write them out in standard notation. ONE CORRECT SOLUTION: (-8x^3)^-2 = (-8)^-2*(x^3)^-2 = 1 / (-8)^2 * 1 / (x^3)^2 = 1/64 * 1/x^6 = 1 / (64 x^6). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). ** * R.2.90 (was R.4.36). Express (x^-2 y) / (x y^2) with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = x^-2 * y x y^2 = x^-2-1 * y^1-2 = x^-3 * y^-1 = 1/x^3 y confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (1/x^2 * y) / (x * y^2) = (1/x^2 * y) * 1 / (x * y^2) = y * 1 / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use positive and negative exponents, then in the last step express everything in terms of positive exponents, as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = 4x^-2 y^-1 z^-1 25x^4 y^2 z^-5 = 4/ 25 x^2 yz^-6 = 4 z ^6 25 x^2 y^2 Law of exponents used: (ab)^n=a^n b^n a^m/ a^n= a^(m-n) Exponents cannot be left negative. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Starting with 4x^-2(yz)^-1/ [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^1 * z^-1: 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5} Grouping the numbers, and the x, the y and the z expression: (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents: (4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further: (4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents: 4z^4/ (25x^6 * y^3 ) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I canceled out the exponents instead of making them positive and subtracting them. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2.122 (was R.4.72). Express 0.00421 in scientific notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4.21x10^-3 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9700 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * R.2.152 \ 150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a. | 97 - 98.6 | > 1.5 | -1.6 | > 1.5 1.6>1.5 b. | 100 - 98.6 | >1.5 | 1.4 | >1.5 1.4<1.5 This is not an unhealthy temperature. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** You can show that T=97 is unhealthy by substituting 97 for T to get | -1.6| > 1.5, equivalent to the true statement 1.6>1.5. But you can't show that T=100 is unhealthy, when you sustitute for T then it becomes | 100 - 98.6 | > 1.5, or | 1.4 | > 1.5, giving us 1.4>1.5, which is an untrue statement. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok