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IRINA_888 [86]
3 years ago
7

Evaluate the expression when x=-5. x²+9x-6

Mathematics
2 answers:
spin [16.1K]3 years ago
8 0

Answer: 64

Step-by-step explanation:

X2+9x-6

When x=-5

(-5)^2+9(5)-6

25+45-6

70-6

64

ch4aika [34]3 years ago
7 0

Answer:

-26

Step-by-step explanation:

{ - 5}^{2}  + 9( - 5) - 6 \\ 25 - 45 - 6 \\  - 26

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Consider the differential equation
Ainat [17]

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W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x

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Differentiate with respect to x

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So, y=e^{\frac{x}{2}} is the solution of the given equation.

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Differentiate with respect to x

y'=e^{\frac{x}{2}}+\frac{x}{2}e^{\frac{x}{2}}=e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )

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Put values of y,y',y'' in 4y''-4y'+y=0

4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}  \right )-4\left (  e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0

To find: W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )

Solution:

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W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x

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