course Mth 158 * R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: * * ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to (x+2)/[(x-2)(x^2+4)]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have two like terms so we write (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We have two like terms so we write (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.52 (was R.7.48). Show how you found and simplified the expression (x - 1) / x^3 + x / (x^2 + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)].Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x-3, x^3+3x and x^3-9x factor intox-3, x(x^2+3) and x(x^2-9) then into(x-3) , x(x^2+3) , x(x-3)(x+3).The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: LCM = x(x-3)(x+3)(x^2+3) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: LCM = x(x-3)(x+3)(x^2+3) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: ********************************************* Question: * R.7.64 (was R.7.60). Show how you found and simplified the difference 3x / (x-1) - (x - 4) / (x^2 - 2x + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2.To express the given expression in terms of the common denominator we then multiply the first expression by (x-1) / (x-1) to get3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2,which gives us 93x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2. To express the given expression in terms of the common denominator we then multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):none ------------------------------------------------ Self-critique Rating: QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to grab onto the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get "