R2 assignment

course MTH 158

06/28 11:16pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your 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.

002. `* 2

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Question: * R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explain how you got your result.

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Your solution:

(2x-3)/y=(2(-2)-3)/3=(-4-3)/3= -7/3

confidence rating #$&* 3

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Given Solution:

* * ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get

(2*(-2) - 3)/3 =

(-4-3)/3=

-7/3. **

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Self-critique (if necessary):

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Self-critique rating #$&* OK

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Question: * R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explain how you got your result.

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Your solution:

||4x|-|5y|=||4*3|-|5*-2|=|12-|-10|=2

confidence rating #$&* 3

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Given Solution:

* * ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get

| | 4*3 | - | 5*-2 | | =

| | 12 | - | -10 | | =

| 12-10 | =

| 2 | =

2. **

* R.2.64 (was R.2.54) Explain what values, if any, cannot be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)

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Your solution:

0 cannot be present because if you divide by zero the answer is zero.

confidence rating #$&* 3

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Given Solution:

* * ** The denominator of this expression cannot be zero, since division by zero is undefined.

Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 is, and only if, either x^2 + 1 = 0 or x = 0.

Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. **

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Self-critique (if necessary):

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Self-critique rating #$&*OK

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Question:

* R.2.76 \ 73 (was R.4.6). What is -4^-2 and how did you use the laws of exponents to get your result?

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Your solution:

-4^-2= -1/16

confidence rating #$&* 3

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Given Solution:

* * ** order of operations implies exponentiation before multiplication; the - in front of the 4 is not part of the 4 but is an implicit multiplication by -1. Thus only 4 is raised to the -2 power.

Starting with the expression -4^(-2):

Since a^-b = 1 / (a^b), we have

4^-2 = 1 / (4)^2 = 1 / 16.

The - in front then gives us -4^(-2) = - ( 1/ 16) = -1/16.

If the intent was to take -4 to the -2 power the expression would have been written (-4)^(-2).**

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Self-critique (if necessary):

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Self-critique rating #$&*OK

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Question:

* Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?

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Your solution:

(3^-2*5^3)/(3^2*5)

1/9*125/9*5=13.8888/45

1/9*125/(9*5) would have worked.

1/9 * 125 = 125 / 9.

Then dividing this by 9 * 5 gives us

(125 / 9) * (1 / 45) = 125 / 405, which reduces to 25 / 81.

It's more instructive (and in the long run easier) to keep things in exponential form, though, and take the powers at the end:

(3^-2*5^3)/(3^2*5) =

(1/3^2 * 5^3) / (3^2 * 5) =

(5^3 / 3^2) / (3^2 * 5) =

5^3 / (3^2 * 3^2 * 5) =

5^2 / (3^4) =

25 / 81.

confidence rating #$&* -2

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Given Solution:

Starting with (3^(-2)*5^3)/(3^2*5):

Grouping factors with like bases we have

3^(-2)/3^2 * 5^3 / 5. Using the fact that a^b / a^c = a^(b-c) we get

3^(-2 -2) * 5^(3-1), which gives us

3^-4 * 5^2. Using a^(-b) = 1 / a^b we get

(1/3^4) * 5^2. Simplifying we have

(1/81) * 25 = 25/81. **

STUDENT QUESTION:

I do not understand how we can ungroup the (3^(-2) *5^3).

INSTRUCTOR RESPONSE

Hopefully this will clarify that operation:

(a / c) * (b / d) = (a * b) / (c * d), since you multiply the numerators to get the numerator and the denominators to get the denominator.

So it must be true that

(a * b) / (c * d) = (a / c) * (b / d).

Now substitute a = 3^(-2), b = 5^3, c = 3^2 and d = 5. You find that

(3^(-2)*5^3)/(3^2*5) = 3^(-2)/3^2 * 5^3 / 5.

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Self-critique (if necessary):

I am completely lost, I tried to do the negative exponent which was wrong. The when I read the solution, I am more confused, I don’t understand the regrouping process. I tried watching the CD on this section and there was nothing about it. Instead of continuing to stress about this I am going to attempt to move on and keep reading about this type problem and hopefully will begin to understand.

Good idea. Don't let yourself get bogged down, but do self-critique and/or submit a question (both of which you did).

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Self-critique rating #$&* Not ok

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Question:

* R.2.94. Express [ 5 x^-2 / (6 y^-2) ] ^ -3 with only positive exponents and explain how you used the laws of exponents to get your result.

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Your solution:

[5x^-2/(6y^-2)]^-3=(6y^2/5x^2)^-3=

confidence rating #$&* -2

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Given Solution:

[ 5 x^-2 / (6 y^-2) ] ^ -3 = (5 x^-2)^-3 / (6 y^-2)^-3, since (a/b)^c = a^c / b^c. This simplifies to

5^-3 (x^-2)^-3 / [ 6^-3 (y^-2)^-3 ] since (ab)^c = a^c b^c. Then since (a^b)^c = a^(bc) we have

5^-3 x^6 / [ 6^-3 y^6 ] . We rearrange this to get the result

6^3 x^6 / (5^3 y^6), since a^-b = 1 / a^b.

STUDENT QUESTION:

I do not see how you can take and seperate the problem down like this has it seems to just have reversed the problem

around in a different ordering and I do not see how this changed the exponets from being negative

Is there anyway you can explain this problem in a little more depth

INSTRUCTOR RESPONSE:

A fundamental law of exponents is that exponentiation distributes over multiplication, so that

(a * b) ^ c = a^c * b^c and

(a / b) ^ c = a^c / b^c

More specifically, if c = -3 then we have

( a * b ) ^ (-3) = a * (-3) * b^(-3) and

( a / b ) ^ (-3) = a ^ (-3) / b^(-3).

Now

a ^ (3) / b^(3) = 1 / a ^ (3) / (1 / b^(3)) and

1 / a ^ (3) / (1 / b^(3)) = 1 / a^3 * (b^3 / 1) = b^3 / a^3.

This principle applies to any string of multiplcations and division, so for example

( a * b / (c * d) ) ^ e = a^e * b^e / (c^e * d^e).

If e = -3 then we would have

( a * b / (c * d) ) ^ (-3) = a^(-3) * b^(-3) / (c^(-3) * d^(-3)).

Since the -3 power is the reciprocal of the 3 power this expression becomes

1/a^(3) * (1/b^(3)) / (1/c^(3) * (1/d^(3))), which is easily seen to be equal to

1 / (a^3 * b^3) / (1 / (c^3 * d^3) ).

Dividing by (1 / (c^3 * d^3) ) is the same as multiplying by (c^3 * d^3) / 1 so

1 / (a^3 * b^3) / (1 / (c^3 * d^3) ) = 1 / (a^3 * b^3) * (c^3 * d^3) = (c^3 * d^3) / (a^3 * b^3).

You should have written the above expressions, which are difficult to read in this notation, on paper, applying the order of operations. The expressions you wrote down should look like the ones below. Be sure you understand the translation from the 'typewriter notation' above to the standard notation depicted below, and be sure you know how to write each of the expressions depicted below in standard notation:

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Self-critique (if necessary):

I got the part about the -2 exponents but did not take care of the -3 exponent. After reading the solution, I think I have a slightly better understanding of what I need to do. I would like to have more practice with this type problem because I am still very confused.

Your book has a number of odd-numbered problems with answers in the back. It's an excellent exercise to attempt these problems and see if you can match the back of the book.

You can always submit questions about problems that aren't covered in the Query, using the Submit Question Form.

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Self-critique rating #$&* Not ok

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Question:

* Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.

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Your solution:

(-8x^3)^-2= -8x^-6=1/8x^6

I combined the exponents and then made them positive by turning them into a fraction. After reading the solution I see that I should have treated 8 and x as two separate units.

confidence rating #$&* -3

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Given Solution:

* * ** ERRONEOUS STUDENT SOLUTION:

(-8x^3)^-2

-1/(-8^2 * x^3+2)

1/64x^5

INSTRUCTOR COMMENT:

1/64x^5 means 1 / 64 * x^5 = x^5 / 64. This is not what you meant but it is the only correct interpretation of what you wrote.

Also it's not x^3 * x^2, which would be x^5, but (x^3)^2.

There are several ways to get the solution. Two ways are shown below. They make more sense if you write them out in standard notation.

ONE CORRECT SOLUTION:

(-8x^3)^-2 =

(-8)^-2*(x^3)^-2 =

1 / (-8)^2 * 1 / (x^3)^2 =

1/64 * 1/x^6 =

1 / (64 x^6).

Alternatively

(-8 x^3)^-2 =

1 / [ (-8 x^3)^2] =

1 / [ (-8)^2 (x^3)^2 ] =

1 / ( 64 x^6 ). **

* R.2.90 (was R.4.36). Express (x^-2 y) / (x y^2) with only positive exponents and explain how you used the laws of exponents to get your result.

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Your solution:

(x^-2y)/(xy^2)=(1/x^2*y)/xy^2 I don’t know how you got from here to the solution given.

I have read the solution and tried rewriting it to get a clearer idea of the steps and I am still very confused

confidence rating #$&* -2

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Given Solution:

(1/x^2 * y) / (x * y^2)

= (1/x^2 * y) * 1 / (x * y^2)

= y * 1 / ( x^2 * x * y^2)

= y / (x^3 y^2)

= 1 / (x^3 y).

Alternatively, or as a check, you could use positive and negative exponents, then in the last step express everything in terms of positive exponents, as follows:

(x^-2y)/(xy^2)

= x^-2 * y * x^-1 * y^-2

= x^(-2 - 1) * y^(1 - 2)

= x^-3 y^-1

= 1 / (x^3 y).

STUDENT QUESTION

I wrote it down on paper and I am still a little confused. I understand it down to the 3rd step and then I lose the meaning of the law of exponents.

Why does it change to:

(1/x^2 * y) multiplied by 1/xy^2 the multiplication throws me off.

INSTRUCTOR RESPONSE

(1/x^2 * y) means ( (1/x^2) * y, which is the same as (y / x^2).

So (1/x^2 * y) / (x * y^2) means

(y / x^2) / (x * y^2).

Division by (x * y^2) is the same as multiplication by 1 / (x * y^2) .

So (y / x^2) / (x * y^2) means

(y / x^2) * (1 / (x * y^2)). Multiplying the numerators and denominators of these fractions we have

(y * 1) / (x^2 * x * y^2), which is

y / (x^3 * y^2). Dividing both numerator and denominator by y we have

1 / (x^3 * y).

Let me know if this doesn't help.

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Self-critique (if necessary):

I find the given solution very confusing, even after rewriting it and trying to follow the steps that way. Where can I find more problems that I can practice on?

Let's take the first of the given solutions:

(1/x^2 * y) / (x * y^2)

= (1/x^2 * y) * 1 / (x * y^2)

= y * 1 / ( x^2 * x * y^2)

= y / (x^3 y^2)

= 1 / (x^3 y).

This solution could just as well have been written

(1/x^2 * y) / (x * y^2)

= (1/x^2 * y) / (x * y^2)

= y / ( x^2 * x * y^2)

= y / (x^3 y^2)

= 1 / (x^3 y)

This leaves off that 1 that appears in the second and third steps, which is there fore a reason but is actually unnecessary, and might be a point of confusion.

If this doesn't seem clear to you, can you tell me specifically what you do and do not understand about each step?

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Self-critique rating #$&* Not Ok

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Question:

* Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result.

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Your solution:

I don’t even know where to begin on this one.

confidence rating #$&*

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Given Solution:

* * ** Starting with

4x^-2(yz)^-1 / [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^-1 * z^-1:

4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5] Grouping the numbers, and the x, the y and the z expression:

(4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents:

(4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further:

(4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents:

4z^4/ (25x^6 * y^3 ) **

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Self-critique (if necessary):

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Self-critique rating #$&* Not OK

You're probably trying to see too much at once. I'll try to break it down a little:

Do you see why (yz)^-1 can be written as y^-1 * z^-1?

Do you then see why 4x^-2(yz)^-1 can be written as 4x^-2 * y^-1 * z^-1?

Do you see why (-5)^2 x^4 y^2 z^-5 = 25 x^4 y^2 z^-5?

Do you see why 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5] = (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5)?

Let me know what you do and do not understand about each of these questions.

You may if you wish submit a copy of the given solution, these questions and your responses using the Submit Question Form. Mark your insertions using #### before and after each insertion. Include a copy of this note, which indicates that this is part of Query 2.

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Question:

* R.2.122 (was R.4.72). Express 0.00421 in scientific notation.

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Your solution:

4.21*10^-3=0.00421

confidence rating #$&* 3

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Given Solution:

* * ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **

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Self-critique (if necessary):

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Self-critique rating #$&* OK

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Question:

* R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.

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Your solution:

9.7*10^3=9700

confidence rating #$&* 3

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Given Solution:

* * ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

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Self-critique (if necessary):

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Self-critique rating #$&* Not OK

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Question:

* R.2.152 \ 150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy?

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Your solution:

|97-98.6|_=1.6 1.6>1.5

|100-98.6|=1.4 1.4>1.5 is not a valid term

confidence rating #$&* 3

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Given Solution:

* * ** You can show that T=97 is unhealthy by substituting 97 for T to get | -1.6| > 1.5, equivalent to the true statement 1.6>1.5.

But you can't show that T=100 is unhealthy, when you sustitute for T then it becomes | 100 - 98.6 | > 1.5, or

| 1.4 | > 1.5, giving us

1.4>1.5, which is an untrue statement. **

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Self-critique (if necessary):

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Self-critique rating #$&*

"

&#Your work looks good. See my notes. Let me know if you have any questions. &#

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