#$&* course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
.............................................
Given Solution: * * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this problem I got the answer as the absolute value of 2. I got this by replacing the x and y variable with their respective inserts. So the problem is 4(3)-5(-2). So the absolute value of 4*3 is 12 minus the absolute value of -10 is 10. The absolute value of -10 is 10 because absolute value references the absolute distance from zero on the number line. So 12-10 is the absolute value of 2. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: We know that the denominator can’t be zero and seeing as the only value for x that would create a zero for the denominator, 0 is the value that cannot be present in the domain. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 is, and only if, either x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * R.2.76 \ 73 (was R.4.6). What is -4^-2 and how did you use the laws of exponents to get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To eliminate the negative exponent we create a fraction and move the negative exponent to the bottom (1/-4^2) since a negative multiplied by a negative is a positive the answer would be 1/16. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 made the mistake of bringing 4 and the negative in front to a power of 2 which I understand now that the negative was not included considering there were no parentheses. ------------------------------------------------ Self-critique Rating:2 ********************************************* 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: Using the law of exponents we know that when exponents are on a base being multiplied they are added and when they are on a base being divided they are subtracted, so when subtracting here -2-2 becomes 3^-4 and when subtracting the exponent 1 from three it becomes 5^2 which changes the problem to 3^-4*5^2 but we can’t have negative exponents so it becomes 1/3^4*5^2. This gives us 1/81*25 and then 1/81*25/1=25/81 as the final answer. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 was a little shaky on the previous problem however I got the right answer. I still don’t think my method is clear. ------------------------------------------------ Self-critique Rating:2
.............................................
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): It seems that I took I one step farther and performed the operation of the exponents (5^3, 6^3). However in your solution you did not. Is there a specific reason you didn’t and if so do I need to do the same when answering these problems?????????? ------------------------------------------------ Self-critique Rating:
.............................................
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: To solve this problem I simply moved the negative exponent to the bottom and subtracted the exponents of y. 1/x^2xy giving 1/x^3y. so the answer for this problem is 1/x^3y. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: To solve this problem you have to move several negative exponents around and distribute them aswell. The first thing I did was distribute the negative 1 to y and z in the numerator, then I moved x^-2 to the denominator along with y^-1 and z^-1. Then I moved z^-5 to the numerator. After this step I combined like terms giving me 4z^5/25x^6y^3z^1. Then I subtracted exponents z^5-z^1 giving the final solution 4z^4/25x^6 y^3 confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ** Starting with 4x^-2(yz)^-1/ [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^1 * z^-1: 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5} Grouping the numbers, and the x, the y and the z expression: (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents: (4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further: (4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents: 4z^4/ (25x^6 * y^3 ) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * R.2.122 (was R.4.72). Express 0.00421 in scientific notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0.00421 expressed in scientific notation is 4.21x10^-3 confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 9.7*10^3 in decimal notation is 9,700 confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * R.2.152 \ 150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this problem we take the numerical values given for T and insert them into the formula. So when T=97 the formula becomes the absolute value of 97-98.6>1.5. So we subtract 98.6 from 97 giving us -1.6 which the absolute value is 1.6 and indeed 1.6>1.5. When T=100 the formula is the absolute value of 100-98.6>1.5. So we subtract 98.6 from100 giving us the absolute value of 1.4. However this statement (1.4>1.5) is not true. It would seem that according to the formula when T=100, it is considered healthy. In a real world application of this formula this might be considered as an unsubstantial margin but according to numerical value and mathematics T=100 is healthy. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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