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goldfiish [28.3K]
3 years ago
9

1In the example problem, 2.4 X 10^6 was rewritten as 24 X 10^5. Explain why those expressions are equivalent.

Mathematics
2 answers:
Vsevolod [243]3 years ago
7 0

Answer:

2.4\times 10^6=24\times 10^5

Reason :

When in a decimal number the decimal shifted right the exponent of 10 is decreased by the number of shifting while if it is shifted left the exponent of 10 is increased by the number of shifting,

Why ?

2.4\times 10^6 =\frac{24}{10}\times 10^6=24\times 10^{-1}\times 10^6=24\times 10^{-1+6}=24\times 10^5

( By using a^m = \frac{1}{a^{-m}} and a^m.a^n=a^{m+n} )

USPshnik [31]3 years ago
3 0

Answer:

See explanation

Step-by-step explanation:

Note that

10^6=1,000,000\\ \\10^5=100,000

Then

2.4\times 10^6=2.4\times 1,000,000=2,400,000\\ \\24\times 10^5=24\times 100,000=2,400,000

As you can see, the results are the same, so

2.4\times 10^6=24\times 10^5

This means that two given expressions are equivalent.

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First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

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|

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Next, we perform row operations to obtain row-echelon form.

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⎢

⎣

1

−

1

1

0

5

−

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3

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2

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|

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→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

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12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

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⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

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