query 3

course Mth 271

I also dont understand these fourth root problems such as4th root of ((3x^2y^3)^4)

The fourth root of x^4 is | x |.

For example, if x = -2, then x^4 = 16 and the 4th root of 16 is 2. In this case x is negative and the result is -x, which is | x |.

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assignment #003

003. `query 3

Applied Calculus I

09-16-2007

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16:38:03

0.3.22 (was 0.3.24 simplify z^-3 (3z^4)

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RESPONSE -->

3z^4z^-3= 3z

confidence assessment: 3

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16:38:17

z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **

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RESPONSE -->

yup

self critique assessment: 3

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16:39:24

0.3.28 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3

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RESPONSE -->

1728s^6/ 729s^3= (64/27)s^3

confidence assessment: 3

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16:41:43

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 -->

i dont understand why its 64s^3/27 and not (64/27)s^3

self critique assessment: 1

It's the same both ways, and both ways of expressing it are correct.

In both cases you first cube s. In the first expression you would then multiply by 64 and divide by 27.

In the second expression you would divide 64 by 27 and multiply by s^3.

Same result either way, and either form is acceptable.

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16:47:52

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 -->

(81x^8y^12)^(1/3)= 4.33 x^(8/3)y^4

(54x^7)^(1/3)= 3.78x^(7/3)

confidence assessment: 0

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16:50:22

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 -->

i dont follow

self critique assessment: 0

&#

In your self-critique you need a phrase-by-phrase analysis of the given solution, detailing everything you do and do not understand.

Deconstruct the given solution and explain in detail what you do and do not understand about every part.

. &#

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16:51:07

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 -->

i dont know

confidence assessment: 0

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16:52:01

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 dont follow

self critique assessment: 0

&#Try deconstructing the given solution to identify what you do and do not understand. Just send a copy of the question, your answer, the given solution and the self-critique. &#

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"

Good work on the first problems.

Try to be more specific about what you do and do not understand in the given solutions on the last two.

Note that the solution given to #34 was for the third root; the text might have slipped in a change to the 4th root, which I apparently missed.