<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7B6%7D%7Bx%20%7D%20%20%3D%204" id="TexFormula1"

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3 years ago

title=" \frac{x}{2} + \frac{6}{x } = 4" alt=" \frac{x}{2} + \frac{6}{x } = 4" align="absmiddle" class="latex-formula">
using quadratic equation....help me if you can

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

6 0

Multiply all terms by x

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A two-person relay race consists of eight times around a football field. Juan runs four and one-sixth times and Antonio runs thr

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Answer:

Step-by-step explanation:

To convert mixed number as improper fractions, use this rule:

$A\frac{B}C}=\frac{(A*C)+B}{C}$

<u>For Juan:</u> ( $4\frac{1}{6}$ )

$4\frac{1}{6}=\frac{(4*6)+1}{6}=\frac{25}{6}$

<u>For Antonio:</u> ( $3\frac{2}{3}$ )

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Answer choice A is correct.

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Transforming the ODE yields

$L\left\{y'' - y'\right\} = L\left\{e^{-3t}\right\}$

$(s^2 Y(s) - sy(0) - y'(0)) - (s Y(s) - y(0)) = \dfrac1{s+3}$

$(s^2 - s) Y(s) = \dfrac1{s+3}$

$Y(s) = \dfrac1{(s^2 - s)(s+3)} = \dfrac1{s(s-1)(s+3)}$

Partial fractions:

$Y(s) = \dfrac as + \dfrac b{s-1} + \dfrac c{s+3}$

$\implies 1 = a(s-1)(s+3) + b s(s+3) + c s(s - 1)$

$\implies 1 = -3 a + (2 a + 3 b - c) s + (a + b + c) s^2$

$\implies \begin{cases}-3a=1 \\ 2a+3b-c=0 \\ a+b+c=0\end{cases} \implies a=-\dfrac13, b=\dfrac14, c=\dfrac1{12}$

$\implies Y(s) = -\dfrac13 \times \dfrac1s + \dfrac14 \times \dfrac1{s-1} + \dfrac1{12} \times \dfrac1{s+3}$

Take the inverse transform:

$\boxed{y(t) = -\dfrac13 + \dfrac{e^t}4 + \dfrac{e^{-3t}}{12}}$

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2 years ago

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Answer:

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Step-by-step explanation:

3 * (3/2)^(8 - 1)

3 *(3/2)^7

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2 years ago

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27/250=0.108

so this decimal is non-repeating and terminating.

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Step-by-step explanation:

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