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

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

In May, Liam and Charlie had the same amount of money in their savings accounts. In June, Liam deposited $140 into his account.

romanna [79]

Let's say both had an original amount of "a".

then Liam deposited 140, so it became a + 140.

then Charlie added 5% of "a" to "a", well, 5% of anything is just (5/100)*anything, so 5% of "a" is just (5/100) *a or 0.05a.

then they checked their amounts and both were equal, what the? let's check.

$\bf \stackrel{\textit{Liam}}{a+140}=\stackrel{\textit{Charlie}}{a+0.05a}\implies a+140=1.05a\implies 140=1.05a-a
\\\\\\
140=0.05a\implies \cfrac{140}{0.05}=a\implies \boxed{2800=a}$

7 0

3 years ago