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Shalnov [3]
3 years ago
6

Solve -2x - 5 = 4 for x.

Mathematics
2 answers:
MrRa [10]3 years ago
7 0

Step-by-step explanation:

-2x - 5 = 4

-2x = 9

x= -4.5

hope it helps

joja [24]3 years ago
6 0

Answer:

<u>-2x </u><u>-</u><u> </u><u>5</u><u> </u><u>=</u><u> </u><u>4</u>

<u>-2x </u><u>=</u><u> </u><u>4</u><u> </u><u>-</u><u> </u><u>-</u><u> </u><u>5</u>

<u>-2x </u><u>=</u><u> </u><u>9</u>

<u>x=</u><u> </u><u>9</u><u>/</u><u>2</u>

<u>x </u><u>=</u><u> </u><u>4</u><u>.</u><u>5</u>

<h2><u>I </u><u>HOPE</u><u> IT</u><u> IS</u><u> HELPFUL</u></h2>

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Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

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4 0
3 years ago
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4 years ago
How to solve 5x+6=-4
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8 0
3 years ago
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What are the solutions to the equation 2x^2 + 9x =5
Maksim231197 [3]

Answer:

\displaystyle -5 \ and \ \frac{1}{2}

Step-by-step explanation:

2x^2 + 9x =5

we have to move 5 in the other side with a different sign, to get a 2 grade ecuation

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a=2

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we have to find delta

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and now we have to use the formula to find the root square

\displaystyle \boxed{ X_{1,2}=\frac{-b\pm\sqrt{\Delta} }{2a} }

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