#$&* course Mth 158 9/5 2 002. `* 2
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Given Solution: * * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: plug 3 where x is and -2 for y and get ||4*3| - |5*-2|| that will become ||12| - |-10|| because of the lines, the -10 has to become positive to pull it out—but the minus sign remains also because of the lines (instead of becoming a plus sign like double negatives usually do). So now the equation simplifies to |12-10| which = |2| = 2 confidence rating #$&*: 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: x cannot = 0 or -1 in the denominator (so it probably doesn't for the rest of the equation either) because there is no way to make the equation of the denominator=0 you can (x^3+x) and simplify by dividing it by x and get x(x^2+1) in x^2+1 we see that the only number that can cancel out +1 so that the equation could equal 0 is -1. however x^2 can never = -1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 is, and only if, either x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): these problems hurt my head ------------------------------------------------ Self-critique Rating: 3
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Given Solution: * * ** order of operations implies exponentiation before multiplication; the - in front of the 4 is not part of the 4 but is an implicit multiplication by -1. Thus only 4 is raised to the -2 power. Starting with the expression -4^(-2): Since a^-b = 1 / (a^b), we have 4^-2 = 1 / (4)^2 = 1 / 16. The - in front then gives us -4^(-2) = - ( 1/ 16) = -1/16. If the intent was to take -4 to the -2 power the expression would have been written (-4)^(-2).** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (3^-2 * 5^3) / (3^2 * 5) group with like basis = (3^-2)/(3^2) * (5^3)/5 if a^b / a^c = a^(b-c) then we can say 3^(-2-2) * 5^(3-1) and get 3^-4 * 5^2 and we can make the -4 exponent positive using a^(-b) = 1 / a^b so we have (1/3^4) * 25 1/81 *25 = 25/81 confidence rating #$&*: 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. ** 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): ok ------------------------------------------------ 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: start with the monster [ 5 x^-2 / (6 y^-2) ] ^ -3 and using (a/b)^c= (a^c) / (b^c) we can “simplify” that to (5x^-2)^-3/ (6y^-2)^-3 then using (ab)^c=(a^c)(b^c) we have (5^-3)(x^-2)^3/ (6^-3)(y^-2)^-3 using (ab)^c= a^(bc) we can simplify the problem to (5^-3)(x^6)/ (6^-3)(y^6) using a^-b = 1/a^b we can flip the numbers a little to make it (6^3)(x^6)/(5^3)(y^6) and simplify it to (216x^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): geez ------------------------------------------------ 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: starting with (-8 x^3) ^ -2 we can use (ab)^c=a^c* b^c to make that -8^-2 *(x^3)^-2 then use a^-b= 1/a^b to make that (1/-8^2) * [1/(x^3)^2] which is simplified to (1/64) * (1/x^6) which becomes 1/(64x^6) 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: take (x^-2 y) / (x y^2) and split it with like numbers to make (x^-2)/x) * (y/y^2) then use a^-b= 1/a^b to get (1/x^3) * (y/y^2) and simplify that to (1/x^3) * (1/y) and get 1/(x^3y) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (x^-2 y) / (x y^2) = (1/x^2) * y / (x * y^2) = y / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use positive and negative exponents, then in the last step express everything in terms of positive exponents, as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y). STUDENT QUESTION I wrote it down on paper and I am still a little confused. I understand it down to the 3rd step and then I lose the meaning of the law of exponents. Why does it change to: (1/x^2 * y) multiplied by 1/xy^2 the multiplication throws me off. INSTRUCTOR RESPONSE (1/x^2 * y) means ( (1/x^2) * y, which is the same as (y / x^2). So (1/x^2 * y) / (x * y^2) means (y / x^2) / (x * y^2). Division by (x * y^2) is the same as multiplication by 1 / (x * y^2) . So (y / x^2) / (x * y^2) means (y / x^2) * (1 / (x * y^2)). Multiplying the numerators and denominators of these fractions we have (y * 1) / (x^2 * x * y^2), which is y / (x^3 * y^2). Dividing both numerator and denominator by y we have 1 / (x^3 * y). Let me know if this doesn't help. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] use (ab)^c=a^c*b^c to make (yz)^-1= y^-1*z^-1 and go ahead and square -5 so you get (4 x^-2 y^-1*z^-1) / [ 25 x^4 y^2 z^-5 ] then use a^-b=1/a^b to rearrange the problem until you get (4z^5)/[25(x^2)(x^4)(y^2)(y^1)(z^1)] and simplify that to (4z^4)/(25x^6*y^3) confidence rating #$&*: 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: move the decimal up 3 places and get 4.21*10^-3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9,700 (move decimal down 3 places) confidence rating #$&*: 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: plug in 97 where T is and get |97-98.6|> 1.5 then |-1.6|>1.5 then 1.6>1.5 showing that T=97 is true to being unhealthy however, when you do the same for 100 you get |1.4|>1.5 then 1.4>1.5 and the statement is untrue and cannot be done 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): ok ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!