#$&*
Mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Assignment R.2
** **
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
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 you plug x and y into the equations:
2(-2)-3/3
-4+3/3
= -7/3
Confidence Assessment: 3
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):
Self-critique Rating: 3
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)||
||12|-|-10||
|2|>> =2
Confidence Assessment: 3
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:
The value of zero (c) is not able to be present in the domain of the expression. The denominator cannot be zero because dividing by zero is undefined. The rest of the values equal out to have a real number as its denominator.
Confidence Assessment: 3
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):
Self-critique Rating: 3
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:
-4^-2
1/-4^2
1/16
I used the laws of exponents by making the exponent into a fraction to get rid of the negative exponent. This resulted in the answer being 1/16.
Confidence Assessment: 3
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):
Self-critique Rating: 3
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 * 5^3)/ (3^2 * 5)
Combine the exponents with like bases by subtracting the exponents
3^-4 * 5^2
Make into a fraction to get rid of the negative exponent
1/3^4 * 5^2
1/81 * 25/1
25/81
Confidence Assessment:
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.
STUDENT SOLUTION (with error)
(3^-2*5^3)/(3^2*5) = 1/9*125/9*5=13.8888/45
INSTRUCTOR CRITIQUE
You almost had it, but you left off the grouping of the denominator.
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.
Self-critique (if necessary):
Self-critique Rating: 3
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.
Your solution:
[ 5 x^-2 / (6 y^-2) ] ^ -3
(5x^-2)^-3 / (6y^-2)^-3
Multiply the exponents into the parentheses
(5^-3*x^6) / (6^-3*y^6)
(1/5^3*x^6) / (1/6^3*y^6)
(1/125*x^6) / (1/216*y^6)
Confidence Assessment: 1
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:
Self-critique (if necessary):
I followed you up until the negative exponents towards the end of the problem. I don’t understand how you got rid of them and could not figure it out in the explanation with a,b, & c. I thought you had to take a negative exponent and put it into a fraction to get rid of the negative. How does changing the order of a, b, & c get rid of the negative?
Self-critique Rating: 1
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:
(-8x^3)^-2
-8x^-6
1/-8x^6
Confidence Assessment: 2
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 y)/ (x y^2)
x^-3 * y^-1
1/x^3*y
Confidence Assessment: 1
Given Solution:
(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).
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.
Self-critique (if necessary):
Self-critique Rating: 3
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:
4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ]
4/25x^-6*y^-1* z^4
(1/ 4/25x^6* y)*z^4
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):
I understand how to simplify exponents by subtracting them if they have like bases but I still do not understand the regrouping. I’m finding it extremely difficult to grasp the concept when I don’t know how to know which is a, b, c, or d.
Self-critique Rating:
Question:
* R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
Your solution:
4.21 x 10^-3
Confidence Assessment: 3
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):
Self-critique Rating:3
Question:
* R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
Your solution:
.0097
Confidence Assessment: 3
Given Solution:
* * ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
Self-critique (if necessary): I solved the problem for if the exponent was negative. I’m having some difficulty remembering which way the decimal moves.
Self-critique Rating: 3
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:
|97-98.6| < 1.5
|-1.6|<1.5
1.6>1.5
|100-98.6| > 1.5
|1.4|>1.5
Confidence Assessment: 3
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):
Self-critique Rating: 3
** **
I do not understand the idea of regrouping the problem to get rid of the negative exponent. I understood the regrouping when you used your example at the beginning of the assignment: (a*b)/(c*d)= (a*b)/(c*d) and you substituted 3^-2 for a, 5^3 for b, 3^2 for c, and 5 for d. I got lost when I had to figure out what a b c and d were. How am I supposed to know which is which and sometimes one of the letters is an exponent? On top of that I did not follow how the negative exponents went away when you changed the grouping. I understand you have explained this thoroughly in the assignment and I apologize if this is plain as day but I am having difficulty.
Thanks in advance!
** **
@&
You've articulated your question very well.
(3^-2 * 5^3) / (3^2 * 5)
is of the form
(a * b) / (c * d)
with a = 3^(-2), b = 5^3, c = 3^2, d = 5.
So
(3^-2 * 5^3) / (3^2 * 5) = (3^-2 / 3^2) * (5^3 / 5)
= 3^-4 * 5^2
= 5^2 / 3^4
= 25 / 81.
In this solution we've used the fact that 3^(-4) = 1 / 3^4.
The purpose of the regrouping was to get the 3's into one expression and the 5's into another so we could combine their powers.
You might well have additional questions about this. If so you need only send a copy of this response, and any relevant parts of your work or the given solutions.
*@
#$&*
Mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Question about problem 37
** **
My question is about number 37 on page 37 of the R.3 section. The question shows a picture of a square with its sides equaling 2 including a shaded circle in the center. It asks you to find the area of the shaded region (which is the circle in the center of the square). I was wondering if you could explain how to complete this problem. I have no idea where to start with these types of problems and always seem to stumble across them.
** **
I understand that the Area of the square is a^2 so it would equal 4. I also understand that the formula for the area of a circle is pi*r^2. You would subtract the area of the square from the area of the circle but how do you find out the radius to be able to do so? Would the radius be 2 since the length of the side of the square is 2?
** **
Thank you!
@&
Good. You've just about got it.
The diameter of the circle would be 2, so its radius would be 1.
From that you can easily find its area.
*@
#$&*
Mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Question about problem 37
** **
My question is about number 37 on page 37 of the R.3 section. The question shows a picture of a square with its sides equaling 2 including a shaded circle in the center. It asks you to find the area of the shaded region (which is the circle in the center of the square). I was wondering if you could explain how to complete this problem. I have no idea where to start with these types of problems and always seem to stumble across them.
** **
I understand that the Area of the square is a^2 so it would equal 4. I also understand that the formula for the area of a circle is pi*r^2. You would subtract the area of the square from the area of the circle but how do you find out the radius to be able to do so? Would the radius be 2 since the length of the side of the square is 2?
** **
Thank you!
@&
Good. You've just about got it.
The diameter of the circle would be 2, so its radius would be 1.
From that you can easily find its area.
*@
After the 'Your Answer' prompt below, insert your answers to the following :
Describe how you constructed your pendulum and out of what (what you used for the mass, its approximate dimensions, what it is made of, what sort of string or thread you used--be as specific as possible).
Describe its motion, including an estimate (you don't have to measure this, just give a ballpark estimate) of how far it swung from side to side and how this distance varied over the time you counted.
Describe what you mean by a 'cycle'. Different people might mean different things, but there are only a couple of reasonable meanings. As long as you describe what you mean we will all understand what you measured.
'Frequency' means the number of cycles in a unit of time. Your counts are frequencies, in cycles/minute. 'Period' means time required for a cycle. Explain how you used your observed frequencies to obtain the periods of the nine pendulums in this experiment.
Your answer (start in the next line):
I constructed my pendulum with a 10k gold ring for the mass, it is roughly 2cm long and I used thin black sewing thread. The pendulum swung at about 12 inches from side to side. The distance varied over time because the length of the string varied.
A complete cycle for me could be described as moving left to right and back to left was 1.
To obtain the frequency of the pendulum I began to swing the pendulum first then I set a timer on my iphone for 1 mintue and then I began to count the number of cycles until the timer went off.
#$&* explanations with std terminology
*#&!
@&
Your tables are now in the right format but they don't answer the given questions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
#$&*
Mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Assignment R.4 Problems
** **
DId you realize that these problems: 111, 114, 117, 120, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 156, 158, 161 for assignment R.4 are not in the new book?
** **
** **
@&
Thanks. I'll have to double-check that.
If they aren't there, of course, you don't need to do them. But there could be other errors in the alignment of the problems with the Queries.
*@