<img src="https://tex.z-dn.net/?f=3.4%20%5Ctimes%204.37" id="TexFormula1" title="3.4 \times 4.37" alt="3.4 \times 4.37" align="a

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serious [3.7K]

3 years ago

bsmiddle" class="latex-formula">

Mathematics

2 answers:

shusha [124]3 years ago

8 0

Hello!

Answer: =14.858

Explanation:

Multiply by the numbers. Product it can also multiply by the numbers, fractions, decimals, factors, and multiples.

3.4 it has one decimal place.

4.37 it has two decimal place.

So, the answer it should be have a three decimal places.

$3.4*4.37=14.858$

Hope this helps!

Thanks :)

-Charlie

balu736 [363]3 years ago

7 0

Multiply it out

4.37

x 3.4

---------

1748

+1311x

--------

14858

Now count how many times you have to move the decimal out on the top to make them whole numbers (3)

So now move 3 decimal places in on your final answer

14858 ==> 14.858

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Step-by-step explanation:

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Answer:

The particular integral of given differential equation

$y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

General solution of given differential equation

$y = y_{c} + y_{p}$

$Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$

Step-by-step explanation:

Step(i):-

Given Differential equation y'' − 5 y' + 4 y = x

Given equation in operator form

D²y - 5 Dy + 4 y = x

⇒ ( D² - 5 D + 4 ) y =x

⇒ f(D) y = Q

where f(D) = D² - 5 D + 4 and Q(x) = x

The auxiliary equation f(m) =0

m²-5 m + 4 =0

m² - 4 m - m + 4 =0

m ( m -4 ) -1 ( m-4) =0

(m - 1) =0 and ( m-4) =0

m = 1 and m =4

The complementary function

$Y_{c} = C_{1} e^{x} + C_{2} e^{4x}$

Step(ii):-

particular integral

Particular integral

$y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2} - 5 D + 4} X$

taking common '4'

$= \frac{1}{4(1 + (\frac{D^{2} - 5 D}{4} ))} X$

$=\frac{1}{4} (1 + (\frac{D^{2} -5D}{4})^{-1} )} X$

applying binomial expression

( 1 + x )⁻¹ = 1 - x + x² - x³ +.....

$=\frac{1}{4} (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X$

Now simplifying and we will use notation D = $\frac{dy}{dx}$

$=\frac{1}{4} (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}$

Higher degree terms are neglected

$=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)$

The particular integral of given differential equation

$y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

Final answer:-

General solution of given differential equation

$y = y_{c} + y_{p}$

$Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$

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horrorfan [7]

Answer

m = -8

(if it's 2 sub n-1; otherwise, likely -10?)

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