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4 years ago

5

A business purchases a computer system for $3000. If the value of the system decreases at a rate or 15% per year, how much is th

e computer worth after a year?

1 answer:

worty [1.4K]4 years ago

6 0

$2550 because 15% of 3000 is 450, so 3000 - 450 = 2550

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Answer: 15/7

Step-by-step explanation: Simplify 6/2 to 3

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4 years ago

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H(n) = n + 2 g(n) = 2n + 1 Find h(n) − g(n)

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3 years ago

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Anika [276]

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$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

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$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$

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$y''(t)+ 8y'(t)-20y(t)=0$

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$x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2$

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