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3 years ago

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Can someone help me with these geometry proofs

1 answer:

White raven [17]3 years ago

6 0

Answer:

See the attachment

Step-by-step explanation:

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The answer to this question is:

<span>Which step comes last when using the divide-center method to find the line on a scatter plot?
</span>B-"Connect the two centers to produce the trend line"

Hoped This Helped, Lauri112311
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• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

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4 years ago

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bearhunter [10]

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3 years ago

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The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em> </em> $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$ <em></em>

<em> General solution of given differential equation</em>

<em> </em> $y = y_{c} + y_{p}$ <em></em>

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

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

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

<em>The auxiliary equation f(m) =0</em>

<em> m²-5 m + 4 =0</em>

m² - 4 m - m + 4 =0

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

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

<em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em> $Y_{c} = C_{1} e^{x} + C_{2} e^{4x}$ <em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

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

<em>taking common '4' </em>

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

<em> </em>

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

<em>applying binomial expression</em>

<em> ( 1 + x )⁻¹ = 1 - x + x² - x³ +..... </em>

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

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

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

<em>Higher degree terms are neglected</em>

<em> </em> $=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)$ <em></em>

<em>The particular integral of given differential equation</em>

<em> </em> $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$ <em></em>

<u><em>Final answer</em></u><em>:-</em>

<em> General solution of given differential equation</em>

<em> </em> $y = y_{c} + y_{p}$ <em></em>

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

<em></em>

<em></em>

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4 0

3 years ago