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Setler [38]
3 years ago
7

What is 0.000001 written as a power of ten?

Mathematics
1 answer:
AleksandrR [38]3 years ago
7 0
a^{-n}=\dfrac{1}{a^n}\\\\10^{-n}=\dfrac{1}{10^n}=0.\underbrace{000...1}_{n}\\\\\text{examples:}\\\\10^{-2}=\dfrac{1}{10^2}=\dfrac{1}{100}=0.\underbrace{01}_{2} \\10^{-4}=\dfrac{1}{10^4}=\dfrac{1}{10,000}=0.\underbrace{0001}_{4}\\\\\text{therefore}\\\\0.\underbrace{000001}_{6}=10^{-6}\\\\\text{Answer:}\ \boxed{B.\ 10^{-6}}
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That question is accompanied by these answer choices:

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Accuracy refers to the closeness of the measure to the real value, while precision, in this case, refers to the level of significant figures that the sacle report.

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Natasha_Volkova [10]

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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

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Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

The characteristic equation

r^2 - 3r + 2 = (r - 1) (r - 2) = 0

has roots at r=1 and r=2. Then the characteristic solution is

y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

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For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

6 0
1 year ago
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