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3 years ago

Use standard notation to write this number. 4.045x10^-3

Mathematics

2 answers:

Inessa05 [86]3 years ago

5 0

<span>4.045 x10^-3 in standard notation = 4,045</span>

bezimeni [28]3 years ago

3 0

In standard notation, the number 4.045x10^-3 can be written as :
0.004045.

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Alex17521 [72]

Answer:

61.66

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

a 2014 mustang originally costs $25000.00 and depreciates at a rate of 8% per year. What is the cost of the car after 7 years?

frutty [35]

$11,000 will bet the cost in 7 years

Given:

Original cost: $25,000

Depreciation rate: 8%

Term: 7 years

Formula for Depreciation:

A = C ( 1 - ( r ) (t) )

A = Future Value

C = Original Cost

r = rate

t = term

Solution:

Substitute the given values to the formula for depreciation.

A = $25,000( 1 - ( 0.08)(7))

A = $25,000( 1 - .56 )

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A = $11,000

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2 years ago

How to make change for a dollar using exactly 50 coins.<br> Is there more than one way?

mylen [45]

Answer:

Conversations of currencies

Step-by-step explanation:

I am 100% sure that's the other way

3 0

3 years ago

A whole pie is cut into 8 equal slices. Three pieces are served. What fraction of the pie is left?

Maksim231197 [3]

1 pie is cut into eighths or 1/8 a slice. 3 are taken out.
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4 0

3 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago