#$&* course Mth 158 your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4(3)-5(2) 12-10 = 2 confidence rating #$&*: 3, I got the same answer as you but I took my problem out of the bars first, is this a problem or will I need to keep them in the bars until the end of the problem. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: I do not understand what is meant by the “domain of the expression. I attempted the problem and a. x=3; b. x=-11/4; c. x=-2/0, d. x=-11/2, I would chose “c” as the value that cannot be present in the domain??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): When completing these problems on the test, since the denominator cannot be zero, do you just work out the denominator part of the problem? ------------------------------------------------ Self-critique Rating:
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: I have really struggled with negative exponents, but it is slowly sinking in. When you have a negative exponent you have to divide to make the exponents positive, you make it into a fraction, in this problem 3^-2, is 1/3^2, then perform the math. 1/3^2 *5^3 = 1/9*125, the denominator is 45. Question: can you reduce this before multiplying ex 5^3 is 5*5*5 and on the bottom it is 3*3*5 if you reduce the 5’s it makes the problem 1/9*25/3*3, so a final answer would be (25/9)/9? I just looked at the given solution. I see what you did but I think I am going to have to work on this one. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): of all of the YouTube videos I have watched, none of them kept it in exponential form till the end, I will try that and work on some extra problems. ------------------------------------------------ Self-critique Rating:
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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): I did get the problem correct but is there a reason that you don’t go ahead and perform the exponent math? ------------------------------------------------ Self-critique Rating:
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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: divided the negative exponent giving me y/x^2+x*y^2, then adding the like terms 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 was confused about where the y^2 went, but I see in an explanation. ------------------------------------------------ Self-critique Rating: ********************************************* 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: still having problems with negative exponents, anything else I can do to help me. I have read over the notes in the book, but I do not feel it explains it enough for me to grasp the concept. I have also watched some YouTube videos but maybe I should review some more. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.2.122 (was R.4.72). Express 0.00421 in scientific notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4.21 x 10^-3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9700 confidence rating #$&*: 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: ********************************************* 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: 97-98.6 = -1.6, -1.6>1.5, 97 is healthy 100-98.6 = 1.4, 1.4>1.5 not healthy 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): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!