#$&* course Mth 158 6/18/13 1:32pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The problem becomes || 4 * 3 | - | 5 * (-2) || = || 12 | - | -10 || = (the absolute value of -10 is simply 10) | 12 - 10 | = | 2 | = 2 confidence rating #$&*: 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: The value of the denominator (x^3 + x) cannot be equal to 0. To find what values of x would make the denominator equal to x I set up an equation and then plugged in the given possible values of x to eliminate the one(s) that x cannot be equal to in this expression. I knew it could not be (a) x = 3 or (b) x = 1 because they would result in x^3 + x = > 0 and (d) x = -1 would result in < 0; all of these are present in the domain of the expression. Just looking at the problem, I feel that (c) x = 0 will be the answer, but I will write it out to double check. x^3 + x using (c) x = 0 0^3 + 0 = 0 + 0 = 0 This shows that in the expression (-9x^2 - x + 1) / (x^3 + x) x is not 0 and therefore not present within the domain. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 is, and only if, either x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Factoring down the problem like this so that x was extracted was a simpler way to do the work. I did not think to do it that way when I worked it, but will try to think of this in the future. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.2.76 \ 73 (was R.4.6). What is -4^-2 and how did you use the laws of exponents to get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -4^-2 = 1/(-4^2) = because the original exponent was negative, I was able to change the sign after dividing 1 by the base 1/-16 The negative sign remained because there were no grouping symbols dictating that it be squared in conjunction with the 4; without the grouping symbols, the problem was basically saying 1/( -1 * 4^2 ) so only the 4 was raised confidence rating #$&*: 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): OK ------------------------------------------------ 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: Because of the negative exponent, the 3^-2 is divided by the 5^3 instead of multiplied; this fraction is still grouped as the numerator of the 3^2 * 5 and must be worked out from that point. ( 5^3 / 3^2 ) / (3^2 * 5) = We can then convert this into a multiplication problem by flipping the denominator (the exponentiation does not need to be worked out quite yet either, as some of it may be canceled out in the next steps) (5^3 / 3^2) * (3^2 *5 / 1) = the 3^2's are canceled out (5^3 / 1 ) * (5 / 1) = 5 * 5 * 5 * 5 = 625 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION: I do not understand how we can ungroup the (3^(-2) *5^3). INSTRUCTOR RESPONSE Hopefully this will clarify that operation: (a / c) * (b / d) = (a * b) / (c * d), since you multiply the numerators to get the numerator and the denominators to get the denominator. So it must be true that (a * b) / (c * d) = (a / c) * (b / d). Now substitute a = 3^(-2), b = 5^3, c = 3^2 and d = 5. You find that (3^(-2)*5^3)/(3^2*5) = 3^(-2)/3^2 * 5^3 / 5. STUDENT SOLUTION (with error) (3^-2*5^3)/(3^2*5) = 1/9*125/9*5=13.8888/45 INSTRUCTOR CRITIQUE You almost had it, but you left off the grouping of the denominator. 1/9*125/(9*5) would have worked. 1/9 * 125 = 125 / 9. Then dividing this by 9 * 5 gives us (125 / 9) * (1 / 45) = 125 / 405, which reduces to 25 / 81. It's more instructive (and in the long run easier) to keep things in exponential form, though, and take the powers at the end: (3^-2*5^3)/(3^2*5) = (1/3^2 * 5^3) / (3^2 * 5) = (5^3 / 3^2) / (3^2 * 5) = 5^3 / (3^2 * 3^2 * 5) = 5^2 / (3^4) = 25 / 81. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had at first worked it out this way, but confused myself into thinking that I could not move the 3^-2 to the denominator so simply. I think I really just overcomplicated it. When I had worked it out the correct way before I rethought it, I had actually arrived at the correct answer. I understand the principle saying the negative exponent makes us move the element, and I am not sure why I confused myself. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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: [ 5 x^-2 / (6 y^-2) ] ^ -3 = [ 5 y^2 / (6x^2) ] ^ -3 = because everything is being raised to a negative power, I will just flip the whole problem over 6^3 * x^6 / (5^3 * y^6) = 276x^6 / 125y^6 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): Ok though I do have a question- ??? When I answered, I simplified the 6^3 and 5^3. Is this not necessary? I thought that we were supposed to simplify it down as much as possible. How do I know if it's enough? ------------------------------------------------ Self-critique Rating: OK
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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 ). ** Self Critique: I left out the negative from what I raised to the exponent. Because of the grouping symbols, I should have raised it to the power even if it was not with the 8 as a -8 and was separated as a -1; squaring it would have resulted in a positive 1 * 64. * 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) = y / (x^2 * xy^2) = y / (x^3 y^2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (x^-2 y) / (x y^2) = (1/x^2) * y / (x * y^2) = y / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use positive and negative exponents, then in the last step express everything in terms of positive exponents, as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y). STUDENT QUESTION I wrote it down on paper and I am still a little confused. I understand it down to the 3rd step and then I lose the meaning of the law of exponents. Why does it change to: (1/x^2 * y) multiplied by 1/xy^2 the multiplication throws me off. INSTRUCTOR RESPONSE (1/x^2 * y) means ( (1/x^2) * y, which is the same as (y / x^2). So (1/x^2 * y) / (x * y^2) means (y / x^2) / (x * y^2). Division by (x * y^2) is the same as multiplication by 1 / (x * y^2) . So (y / x^2) / (x * y^2) means (y / x^2) * (1 / (x * y^2)). Multiplying the numerators and denominators of these fractions we have (y * 1) / (x^2 * x * y^2), which is y / (x^3 * y^2). Dividing both numerator and denominator by y we have 1 / (x^3 * y). Let me know if this doesn't help. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to simplify enough and did not cancel out the y's. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] = 4 z^5 / ( 25 x^4 y^2 x^2 y z) = 4 z^4 / (25 x^6 y^3) confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.2.122 (was R.4.72). Express 0.00421 in scientific notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4.21 * 10^-3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9,700 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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: If T = 97 | 97 - 98.6 | > 1.5 | -1.6 | > 1.5 1.6 > 1.5 This is a true statement, and shows that a temperature of 97 is unhealthy If T = 100 | 100 - 98.6| > 1.5 | 1.4 | > 1.5 1.4 > 1.5 This is an incorrect statement, and shows that a temperature of 100 is unhealthy confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!