course Mth 158 {жpassignment #002
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11:28:54 R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.
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RESPONSE --> First I replaced x as (-2). (2(-2)-3). then replaced y as 3. (2(-2) - 3)/ 3. Then I used the order of operations starting with the parentheses. (-4 - 3) / 3. Then I simplified the parentheses. I got -7 / 3 as my answer. confidence assessment: 2
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11:29:30 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
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RESPONSE --> That is how I did it. I followed the order of operations and got the correct answer. self critique assessment: 3
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11:33:43 R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.
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RESPONSE --> I replaced x and y. |4(3)| - |5(-2)|. Then I worked what is in the absolute value bars. |12| - |-10|. Then I changed the -10 to positive because there cannot be a negative absolute value. |12| - |10| then I got |2| which equals 2. confidence assessment: 3
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11:33:55 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **
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RESPONSE --> I got it right. self critique assessment: 3
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11:35:42 R.2.64 (was R.2.54) Explain what values, if any, must not be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)
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RESPONSE --> 0 must be excluded because it would make the denominator 0. confidence assessment: 1
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11:37:33 ** 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 only if 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. **
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RESPONSE --> That was my reasoning behind my answer. I knew that the denominator could not be 0. self critique assessment: 2
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11:39:28 R.2.73 (was R.4.6). What is (-4)^-2 and how did you use the laws of exponents to get your result?
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RESPONSE --> (-4)^2 is 16. I squared everything in the parentheses including the negative. I know that a negative times a negative equals a positive. confidence assessment: 3
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11:40:43 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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RESPONSE --> I thought that you only moved the number to the bottom of the fraction when the exponent is negative. I'll remember that next time. self critique assessment: 2
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11:45:51 Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?
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RESPONSE --> Since the exponent on the first 3 is negative you must move (3^-2* 5^3) to the bottom of the fraction that gives you (3^2*5^3)(3^2*5). Then you solve the first set of parentheses staring with the exponents. (9* 125) is 1,125. Then you solve the second set starting with the exponent. (9*5) is 45. Then you multiply the two answers together (1,125 *45) =50,625 confidence assessment: 1
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11:48:05 ** (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. **
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RESPONSE --> I did that one completly wrong. I understand that I should've grouped factors with like bases, that makes it much easier. I see how to find the correct answer using the laws of exponents. self critique assessment: 2
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11:53:23 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.
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RESPONSE --> First you must distribute the -3 to everything in parentheses. That gives you (5^-3*x^6) / (6^-3*y^6). To make the exponents positive you must switch the denominator and the numerator. Giving you (6^3y^6)/ (5^3x^6) If you simplify that you find the answer to be (216y^6)/ (125x^6) confidence assessment: 1
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11:54:17 [ 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.
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RESPONSE --> That is what I did and I went ahead and found 6^3=216 and 5^3=125. self critique assessment: 2
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11:56:36 Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> Since the exponent is negative you must move everything to the bottom of the fraction. 1/ (-8x^3)^2. Then you square everything in parentheses. 1/ (-64 x^6). confidence assessment: 2
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11:57:54 ** 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^5). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). **
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RESPONSE --> I found the second correct solution. self critique assessment: 3
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12:00:25 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.
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RESPONSE --> Again, you must move the top of the fraction to the bottom since the exponent is negative. That gives you (xy^2)(x^2y)= x^3*y^3 confidence assessment: 1
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12:02:28 ** (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 exponents on term 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).**
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RESPONSE --> I didn't do that one correctly. I moved the entire numerator to the bottom instead of just what was negative. That is where I went wrong. self critique assessment: 2
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12:16:47 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.
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RESPONSE --> first, I moved 4x^-2 ( y z)^-1 to the bottom and z^-5 to the top.
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12:19:14 ** 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 ) **
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RESPONSE --> I still don't have this grouping thing down. I'm sure exactly how to use it. self critique assessment: 2
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12:19:48 R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21*10^-3 confidence assessment: 2
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12:20:02 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **
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RESPONSE --> I got it. self critique assessment: 3
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12:22:55 R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9.700 *10^3=9,7000 confidence assessment: 3
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12:23:03 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> I got it. self critique assessment: 3
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12:26:35 R.2.150 (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?
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RESPONSE --> You put 97 in place of T |97-98.6| >1.5. You solve for absolute value. |1.6| > 1.5. 1.6 is greater than 1.5. Then you put 100 in place of T. |100-98.6| >1.5 Then solve for the absolute value. |1.4| > 1.5. This is not true so a temperature of 100 isn't considered unhealthy. confidence assessment: 2
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12:27:05 ** 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. **
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RESPONSE --> That is how I solved it and I found that 1.4>1.5 is not true. self critique assessment: 2
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