QA 18

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course mth 151

3/24/2015 11:13

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.

018. Base-10 Place-value Number System

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Question: `q001. There are 7 questions in this set.

From lectures and textbook you will learn about some of the counting systems used by past cultures. Various systems enabled people to count objects and to do basic arithmetic, but the base-10 place value system almost universally used today has significant advantages over all these systems.

The key to the base-10 place value system is that each digit in a number tells us how many times a corresponding power of 10 is to be counted.

For example the number 347 tells us that we have seven 1's, 4 ten's and 3 one-hundred's, so 347 means 3 * 100 + 4 * 10 + 7 * 1.

Since 10^2 = 100, 10^1 = 10 and 10^0 = 1, this is also written as

3 * 10^2 + 4 * 10^1 + 7 * 10^0.

How would we write 836 in terms of powers of 10?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

8*10^2+3*10^1+6*10*0

confidence rating #$&*:3

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Given Solution:

836 means 8 * 100 + 3 * 10 + 6 * 1, or 8 * 10^2 + 3 * 10^1 + 6 * 10^0.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q002. How would we write 34,907 in terms of powers of 10?

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Your solution:

3*10000+ 4 *1000+9*100+0*10+7*1

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

34,907 means 3 * 10,000 + 4 * 1000 + 9 * 100 + 0 * 10 + 7 * 1, or 3 * 10^4 + 4 * 10^3 + 9 * 10^2 + 0 * 10 + 7 * 1.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q003. How would we write .00326 in terms of powers of 10?

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Your solution:0 * 10^-1 + 0 * 10^-2 + 3 * 10^-3 + 2 * 10^-4 + 6 * 10^-5

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

First we note that

.1 = 1/10 = 1/10^1 = 10^-1,

.01 = 1/100 = 1/10^2 = 10^-2,

.001 = 1/1000 = 1/10^3 = 10^-3, etc..

Thus .00326 means

0 * .1 + 0 * .01 + 3 * .001 + 2 * .0001 + 6 * .00001 =

0 * 10^-1 + 0 * 10^-2 + 3 * 10^-3 + 2 * 10^-4 + 6 * 10^-5 .

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q004. How would we add 3 * 10^2 + 5 * 10^1 + 7 * 10^0 to 5 * 10^2 + 4 * 10^1 + 2 * 10^0?

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Your solution:

300+50+7 +500+40+2 =899

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

We would write the sum as

(3 * 10^2 + 5 * 10^1 + 7 * 10^0) + (5 * 10^2 + 4 * 10^1 + 2 * 10^0) ,

which we would then rearrange as

(3 * 10^2 + 5 * 10^2) + ( 5 * 10^1 + 4 * 10^1) + ( 7 * 10^0 + 2 * 10^0),

which gives us

8 * 10^2 + 9 * 10^1 + 9 * 10^0. This result would then be written as 899.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q005. How would we add 4 * 10^2 + 7 * 10^1 + 8 * 10^0 to 5 * 10^2 + 6 * 10^1 + 4 * 10^0?

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Your solution:

478+ 564 = 1042

confidence rating #$&*:ok

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

We would write the sum as

(4 * 10^2 + 7 * 10^1 + 8 * 10^0) + (5 * 10^2 + 6 * 10^1 + 4 * 10^0) ,

which we would then rearrange as

(4 * 10^2 + 5 * 10^2) + ( 7 * 10^1 + 6 * 10^1) + ( 8 * 10^0 + 4 * 10^0),

which gives us

9 * 10^2 + 13 * 10^1 + 12 * 10^0.

Since 12 * 10^0 = (2 + 10 ) * 10^0 = 2 * 10^0 + 10^1, we have

9 * 10^2 + 13 * 10^1 + 1 * 10^1 + 2 * 10^0 =

9 * 10^2 + 14 * 10^1 + 2 * 10^0.

Since 14 * 10^1 = 10 * 10^1 + 4 * 10^1 = 10^2 + 4 * 10^1, we have

9 * 10^2 + 1 * 10^2 + 4 * 10^1 + 2 * 10^0 =

10^10^2 + 4 * 10^1 + 2 * 10^0.

Since 10*10^2 = 10^3, we rewrite this as 1 * 10^3 + 0 * 10^2 + 4 * 10^1 + 2 * 10^0.

This number would be expressed as 1042.

STUDENT SOLUTION

(4 x 10^2 + 5 x 10^2) + (7 x 10^1 + 6 + 10^1) + (8 x 10^0 + 4 x 10^0)

adds up to

9 x 10^2 + 13 x 10^1 + 12 x 10^0 = 1042

INSTRUCTOR RESPONSE

You got

9 x 10^2 + 13 x 10^1 + 12 x 10^0 = 1042

But this isn't in its final powers-of-10 notation.

13 * 10^1 isn't a legal expression. Since 13 is greater than 9, you would use the fact that 13 * 10^1 = 10^2 + 3 * 10^1 to write this in correct notation.

Your expression would then become

9 x 10^2 + 10^2 + 3 x 10^1 + 12 x 10^0

Also 12 * 10^0 = 10^1 + 2 * 10^0, so your expression is equivalent to

9 x 10^2 + 1 * 10^2 + 3 x 10^1 + 10^1 + 2 x 10^0

When we add the like powers of 10 we find that 9 * 10^2 + 10^2 = 10 * 10^2, which is 10^3.

Since 3 * 10^1 + 10^1 = 4 * 10^1.

your final expression should be

10^3 + 4 * 10^1 + 2 * 10^0.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q006. Write each of the following in expanded notation using the highest possible powers of 10, and explain your reasoning for each, using only expanded notation in your explanations:

14 * 10^3

4 * 10^4 + 14 * 10^3

8 * 10^3 + 17 * 10^2 + 21 * 10^1 + 15 * 10^0

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Your solution:140 1* 10 ^2 + 4*10 ^ 1 + 0 *10 ^0

180.4 1 * 10 ^ 2 + 8*10^ 1 + 0*10^0 + 4 *10 ^-1

2563 2 * 10^3 + 5 * 10 ^ 2 + 6*10^1+3*10^1

confidence rating #$&*:

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Question: `q007. Show how we would add 8 * 10^2 + 7 * 10^0 to 4 * 10^2 + 5 * 10^0, performing all steps in expanded powers of 10.

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Your solution:

(8*10^2+7*10^0 ) + ( 4*10^2 + 5*10 ^0)

(8*10^2)+(7*10^0 ) + ( 4*10^2) + (5*10 ^0)

(800 + 7) + (40+5)

@&
All steps were to be done in expanded powers of 10.

You have converted in this step to standard base-10 aritmetic, rather than using expanded powers of 10.
*@

(807)+(45)

852

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Self-critique Rating:

ok"

Self-critique (if necessary):

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Self-critique rating:

ok"

Self-critique (if necessary):

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Self-critique rating:

#*&!

QA 18

@& All steps were to be done in expanded powers of 10. You have converted in this step to standard base-10 aritmetic, rather than using expanded powers of 10. *@

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

@&
All steps were to be done in expanded powers of 10.

You have converted in this step to standard base-10 aritmetic, rather than using expanded powers of 10.
*@