course mth271 13 July 2009 @ 2000 hours Mth006Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No 0.3.30 is (3 root of X^2)^3 (12s^2/9s))^3= (4/3S)^3
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Given Solution: `a Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not finish by cubing Self-critique Rating: ********************************************* Question: `q0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No that is not 0.3.38 in my book ((3x^2y^3)^4)^(1/3) ( 81x^6*16y7)^(1/3) (27x^6*8y^6)^(1/3)(3*2y)^(1/3) 3x^2*2y^2*(3*2y)^(1/3) (54*x^7)^(1/3) (27*2*x^6*x)^(1/3) (27x^6)^(1/3) *2x^(1/3) 3x^2*(2x)^(1/3)
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Given Solution: `a To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q0.3.58 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(1+R)(1+(1+R)+(1+R)^2……)
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Given Solution: `a Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: STUDENT QUESTION ON TEXT EXAMPLE, WITH INSERTED INSTRUCTOR RESPONSES: Simplify the expression by factoring. a.) 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2. The next steps I do not understand how they did. Next step: = (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] It might be easier to see how this was done if you multiply the last expression out. (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] = (x+1)^1/2(2x-3)^3/2 * 3(2x-3) + (x+1)^1/2(2x-3)^3/2 * 10(x+1) by the distributive law. (x+1)^1/2(2x-3)^3/2 * 3(2x-3) = 3(x+1)^1/2 (2x - 3)^5/2 and (x+1)^1/2(2x-3)^3/2 * 10(x+1) = 10(x+1)^3/2 (2x-3)^3/2. So (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] = 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2 Now, to factor 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2 you use the same ideas, but in reverse order. We first note than both terms contain powers of x+1 and of 2x - 3. Whichever power of x + 1 is the lesser, that power will be common to both terms of our expression. In this case 1/2 is the common power, so (x + 1)^(1/2) is common to both terms. Whichever power of 2x - 3 is the lesser, that power will be common to both terms of our expression. In this case 3/2 is the common power, so (2x - 3)^(3/2) is common to both terms. So we factor (x + 1)^(1/2) * (2x - 3)^(3/2) out of both terms of the original expression. (To figure out what each remaining term will be, just divide each term by (x + 1)^(1/2) * (2x - 3)^(3/2)). "