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2 years ago

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What is (-47/10)-(53/10) is it -3/5 or -10 the calculator says -10 I say -3/5

2 answers:

horsena [70]2 years ago

8 0

-10. because you already have common denominators, just subtract 53 from -47 (which is the same as adding negative #'s) and you get -100.
-100/10 = -10
Hope that helps:)

Rainbow [258]2 years ago

4 0

$-\frac{47}{10}-\frac{53}{10}=\\
-\frac{100}{10}=\\
-10$

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0.4(x-0.45)=9.2 PLZZZZZ HELP!!!!!!!

Helen [10]

Answer:

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Step-by-step explanation:

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2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

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Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

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$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

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Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

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so the solution is

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Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

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so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

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zubka84 [21]

Answer:

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Step-by-step explanation:

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2 years ago

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ivolga24 [154]

Answer:

66

Step-by-step explanation:

complimentary angles add up to 90 so:

3x-6 + 2x +46 = 90

5x +40 = 90

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2 years ago

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Answer:

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2 years ago