QA 18

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course Mth 151

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 5 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?

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Your solution:

8 x 100 + 3 x 10 + 6 x 1 = 836

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): OK

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self-critique rating #$&*: OK

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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 x 10,000 + 4 x 1000 + 9 x 100 + 0 x 10 + 7 x 1 = 34,907

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): OK

Self-critique Rating: OK

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

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Your solution:

0 x .1 + 0 x .01 + 3 x .001 + 2 x .0001 + 6 x .00001 = .00326

confidence rating #$&*: 2

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

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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): OK

Self-critique Rating: OK

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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:

You would start off breaking them into 2

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

Which would then be rearranged as:

(3 x 10^2 + 5 x 10^2) + (5 x 10^1+ 4 x 10^1) + (5 x 10^2 + 4 x 10^1 + 2 x 10^0)

Which would end up giving you:

8 x 10^2 + 9 x 10^1 + 9 x 10^ 0 = 889

Confidence Assessment: 2

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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): OK

Self-critique Rating: OK

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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:

You would split them into:

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

Which is rearranged as:

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

Which ends up giving you:

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

confidence rating #$&*: 1

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

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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.

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Self-critique (if necessary):

This one confused me!

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 rating #$&*: 2

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

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