Assignment 2 R1

Your work on this assignment is good overall, though you had a few errors and there were things you should have self-critiqued that you didn't. See my notes for more details on self-critique.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

......!!!!!!!!...................................

19:18:06 query R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.

......!!!!!!!!...................................

RESPONSE --> First fill in the numbers into the equation (2(-2)-3) /3 now multiply the 2 and -2 due to order of operations. (-4-3)/3 next subtract the -4-3 to give you 7, and last divide by 3. -2.3repeating

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

......!!!!!!!!...................................

19:18:23 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:21:44 query R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.

......!!!!!!!!...................................

RESPONSE --> Fill in the numbers | |4(3)| - |5(-2)| | now multiply | |12| - |-10| | and because its in absolute values you remove the negatives from the 10. Then Subtract the 12-10 and you get the absolute value of 2 | 12 - 10 | | 2 |

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

......!!!!!!!!...................................

19:21:56 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:25:50 query R.2.64 (was R.2.54) Explain what values, if any, must not be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)

......!!!!!!!!...................................

RESPONSE --> The x will no be represented because it will cancel out with the other x. the problem will be set up as so -9x^2 -x+1/X^3 +x. The x and -x cancel out leaving.. -9x^2+1/x^3 you subtract the x^3 and you get -x^3-9x^2+1, mulitply by -1 to get x^3+9x^2-1

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

......!!!!!!!!...................................

19:26:31 ** 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 only if 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. **

......!!!!!!!!...................................

RESPONSE --> I understand now

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

......!!!!!!!!...................................

19:28:56 query R.2.73 (was R.4.6). What is (-4)^-2 and how did you use the laws of exponents to get your result?

......!!!!!!!!...................................

RESPONSE --> The answer is 1/16 because the exponent is negative you put the 1 over the 4 to give you 1/-4^2...you then multiply 4*4 because its the 2nd power and you get 1?16

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

......!!!!!!!!...................................

19:29:22 query Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?

......!!!!!!!!...................................

RESPONSE --> ok.

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

......!!!!!!!!...................................

19:29:29 ** (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. **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:33:07 query 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.

......!!!!!!!!...................................

RESPONSE --> You put the whole equation under 1? That changes all the negative exponents to positive. 1/ [5x^2/(6y^2)]^3

Good start but this needs to be put into the form of a simple fraction; right now it's a complex fraction (it's a fraction whose denominator is itself a fraction).

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

......!!!!!!!!...................................

19:33:27 [ 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.

......!!!!!!!!...................................

RESPONSE --> ok

You should have self-critiqued this problem, since your solution differed from the given solution.

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

......!!!!!!!!...................................

19:35:34 query Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.

......!!!!!!!!...................................

RESPONSE --> It would be -8x^3*^-2 -8x^-6, but that under one 1/-8x^6

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

......!!!!!!!!...................................

19:36:05 ** 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^5). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:38:31 query 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.

......!!!!!!!!...................................

RESPONSE --> Bring the x^-2 to the bottom to change it to positive you'll have y/x^2*x y now you'll have to add that x in so you have for a final answer y/x^3 y

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

......!!!!!!!!...................................

19:38:42 ** (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 exponents on term 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).**

......!!!!!!!!...................................

RESPONSE --> ok

You were close, but this should also have been self-critiqued.

In a good self-critique you need identify the specific things you do and do not understand in the given solution, and either demonstrate your understanding or ask specific questions about what you don�t understand, so I can help you address what you do not understand.

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

......!!!!!!!!...................................

19:44:17 query 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.

......!!!!!!!!...................................

RESPONSE --> You will have 4x^-2 y^-1 z^-1/ [25 x^4 y^2 z^-5] Take the x^-2 to the bottom, y^-1 to the bottom and z^-1 to the bottom and bring up the z^-5. You now have 4z^5/25x^4*x^2 y^2*y^1 z the laws tell you to add the like variables exponents. Your final answer is 4z^5/25x^6 y^3 z

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

......!!!!!!!!...................................

19:44:48 ** 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 ) **

......!!!!!!!!...................................

RESPONSE --> Ok, i didn't subtract the 1 z

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

......!!!!!!!!...................................

19:45:08 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.

......!!!!!!!!...................................

RESPONSE --> 4.21 x 10^-3

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

......!!!!!!!!...................................

19:45:18 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:45:33 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.

......!!!!!!!!...................................

RESPONSE --> 9700

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

......!!!!!!!!...................................

19:45:39 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

......!!!!!!!!...................................

RESPONSE --> ok

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

......!!!!!!!!...................................

19:47:49 query R.2.150 (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?

......!!!!!!!!...................................

RESPONSE --> Fill each answer into T | 97-98.6|> 1.5 | -1.6 | > 1.5 1.6 > 1.5 which is true |100-98.6|>1.5 |91.4|>1.5 91.4 > 1.5 which is also true

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

......!!!!!!!!...................................

19:48:26 ** 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. **

......!!!!!!!!...................................

RESPONSE --> I subtracted wrong on the 100-98.6

................................................."