assignment 3 query

course 271

6/20

UcƑݓT}xڸwassignment #003

003. `query 3

Applied Calculus I

06-20-2009

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22:53:08

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

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RESPONSE -->

4-3=3z

self critique assessment: 3

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22:57:52

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 -->

simplify to (4/3 s)^3, which is equal to about 2.37s^3

self critique assessment: 3

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23:08:44

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 -->

Factor 54 into 3cbrt2 and x^7 into x^2cbrtx, so we get 3x^2cbrt(2x)

Factor (3x^2 y^3)^4) ^ (1/3) to (3x^2 y^3)cbrt(3x^2 y^3), which can be further simplified into (3x^2 y^4) cbrt(3x^2)

self critique assessment: 3

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23:10:52

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 -->

P(1+r)^n

self critique assessment: 2

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&#Good work. Let me know if you have questions. &#