Assignment 2

course MTH 158

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

*********************************************

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.

*********************************************

Your solution:

First substitute the x=-2 and y=3 into the expression

(2x-3)/y

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

-7/3

Confidence Assessment: 0

.............................................

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 0

Self-critique Rating: 0

*********************************************

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

*********************************************

Your solution:

| | 4*3|-|5*(-2) | | First substitute x=3 and y=-2 in to the expression

|12 -10 |

| 2 |

2

Confidence Assessment: 0

.............................................

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)

*********************************************

Your solution:

Zero cannot be in the denominator because division by zero is not defined.

(-9*0^2-0+1)/(0^3+0)

1/0

Confidence Assessment:

.............................................

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

0

Self-critique Rating: 0

*********************************************

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?

*********************************************

Your solution:

a^-b = 1 / (a^b)

1/4^2

1/16

Confidence Assessment: 0

.............................................

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).**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 0

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

Self-critique Rating: 0

*********************************************

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?

*********************************************

Your solution:

3^-2/3^2*5^3/5

3^(-2-2)*5^(3-1) you are using the law a^b / a^c = a^(b-c)

3^-4*5^2

1/3^4*25 the law a^-b = 1 / (a^b)

1/81*25=1/81*25/1

25/81

Confidence Assessment: 0

.............................................

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. **

* 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.

[ 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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 0

Self-critique Rating: 0

*********************************************

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.

*********************************************

Your solution:

1/[(-8x^3)^2] you use the law a^-b = 1 / (a^b)

1/-8^2*(x^3)^2 you use the law (ab)^n=a^n*b^n

1/64*x^6 you use the law (a^m)^n=a^m*n

Confidence Assessment: 0

.............................................

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.

*********************************************

Your solution:

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

X^-2-1*y^1-2

X^-3y^-1

1/x^3y

Confidence Assessment:

.............................................

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 0

Self-critique Rating: 0

*********************************************

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.

*********************************************

Your solution:

First square the -5^2 in the denominator. You will get:

4x^-2*(yz)-1/25*x^4*y^2*z^-5

4x^-2*y^-1*z^-1/25*x^4*y^2*z^-5 = the law (ab)^n=a^n*b^n

(4/25)*(x^-2/x^-6)*(y^-1/y^2)*(z^-1/z^-5) = the law a^m/a^n = a^m-n and also group like terms together

4z^4/25x^6*y^3 = the law a^-n = 1/a^n

Confidence Assessment: 0

.............................................

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 ) **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 0

Self-critique Rating: 0

*********************************************

Question:

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

*********************************************

Your solution:

4.21*10^-3

Confidence Assessment:0

.............................................

Given Solution:

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):0

Self-critique Rating:0

*********************************************

Question:

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

*********************************************

Your solution:

9.7*1000=9700

Confidence Assessment:0

.............................................

Given Solution:

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):0

Self-critique Rating:0

*********************************************

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?

*********************************************

Your solution:

First substitute the first temperature in the expression T=97

|97-98.6 |>1.5

|-1.6 |>1.5

1.6>1.5 1.6 is greater than 1.5

Then substitute the second temperature in the expression T=100

|100-98.6 |>1.5

|1.4 |>1.5

1.4>1.5 1.4 is not greater than 1.5

Confidence Assessment: 0

.............................................

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): o

Self-critique Rating: o

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

Do review the standards for confidence and self-critique ratings as explained in the documents you submitted for Orientation Part 3.