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Allisa [31]
3 years ago
15

What is 4 over 18 in fractions

Mathematics
1 answer:
vredina [299]3 years ago
5 0
It is 4/18 which can be simplified into 2/9
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Answer:

B

Step-by-step explanation:

g(4) = 8

f(8) = 3(8)^2 - 3(8) 6 = 174

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Find dy/dx for 4 - xy = y^3
storchak [24]

Answer:

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Step-by-step explanation:

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\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}

Solving for dy/dx: Addind x dy/dx both sides of the equation:

-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}

Common factor dy/dx on the right side of the equation:

-y=(3y^2+x)\frac{dy}{dx}

Dividing both sides of the equation by 3y^2+x:

\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}

7 0
3 years ago
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