Assignment 18

course Mth151

tvƒݰ]ۭӥStudent Name:

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assignment #018

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08:58:35

`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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RESPONSE -->

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

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08:58:38

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

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RESPONSE -->

ok

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08:59:44

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

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RESPONSE -->

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

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08:59:47

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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RESPONSE -->

ok

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08:59:53

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

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RESPONSE -->

?

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09:00:23

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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RESPONSE -->

ok...i blanked

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09:02:20

`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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RESPONSE -->

first multiply the numbers by the corresponding powers of ten and then add the products....

300+50+7=357

and

500+40+2=542

add 357 to 542 and the sum is 899

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09:02:26

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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RESPONSE -->

k

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09:03:17

`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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RESPONSE -->

400+70+8=478

500+60+4=564

478+564=1042

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09:03:21

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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RESPONSE -->

ok

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"

Good, but you didn't do those last two problems in the prescribed way, and didn't self-critique. Be sure you understand the expanded notation. Among other things, it's on the test. Let me know if you have questions.