assn 18

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course Mth 151-

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 * 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):

ok

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Self-critique rating #$&*

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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 * 10^4) + (4 * 10^3) + (9 * 10^2) + (0 * 10^1) + (7 * 10^0)

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

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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):

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:

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

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):

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:

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

confidence rating #$&* 3

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

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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):

I didn’t take it far enulf but its just because I wasn’t looking for it. I understand how to do it though

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Self-critique rating #$&*

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