course Mth 271
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18:08:38 0.3.22 (was 0.3.24 simplify z^-3 (3z^4)
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RESPONSE --> z^-3 (3x^4 (3z^4)/z^3 3z confidence assessment: 2
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18:08:49 z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
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RESPONSE --> self critique assessment: 3
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18:11:53 0.3.28 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
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RESPONSE --> ((12x^2)/(9x))^3 (1728x^5)/(729x^3) (576x^2)/(243) (192x^2) (64x^2)/(27) confidence assessment: 2
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18:12:06 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. **
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RESPONSE --> self critique assessment: 3
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18:17:19 0.3.34 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
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RESPONSE --> ( (3x^2 y^3)^4) ^ (1/3) ( (81x^8y^12)^ (1/3) 3x^2y^3 (54 x^7) ^ (1/3) (2x^4) (3x^3) ^(1/3) confidence assessment: 1
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18:20:06 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). **
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RESPONSE --> In the second equation, I factored 54 wrong, ending up with the wrong answer. This too was something I didn't understand completely, but after looking over this solution, I can see how we factor out the prime numbers in the coefficients, and taking the cubes out to come up with the right answers. self critique assessment: 2
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18:20:51 0.3.62 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
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RESPONSE --> A= P (1+r)^n confidence assessment: 1
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18:25:17 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). **
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RESPONSE --> I understand this concept now. When you have a problem like this, you can factor out the like terms, and come up with the factored expression. self critique assessment: 2
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