$\bf x^2+3y^2=12\implies \stackrel{chain~rule}{2x\cfrac{dx}{dt}}+3 \stackrel{chain~rule}{\left(2y\cfrac{dy}{dt}\right)}=0
\\\\\\
2x\cfrac{dx}{dt}+6y\cfrac{dy}{dt}=0
\implies 
6y\cfrac{dy}{dt}=-2x\cfrac{dx}{dt}\implies \cfrac{dy}{dt}=\cfrac{-2x\frac{dx}{dt}}{
6y\frac{dy}{dt}}
\\\\\\
\boxed{\cfrac{dy}{dt}=-\cfrac{x\frac{dx}{dt}}{3y}}\\\\
-------------------------------$

$\bf \textit{now, when y = 1, what is \underline{x}?}\qquad x^2+3y^2=12\implies x^2+3(1)^2=12
\\\\\\
x^2=9\implies x=\sqrt{9}\implies x=3\\\\
-------------------------------\\\\
\begin{cases}
y=1\\
x=3\\
\frac{dx}{dt}=2
\end{cases}\implies \cfrac{dy}{dt}=-\cfrac{3\cdot 2}{3\cdot 1}\implies \cfrac{dy}{dt}=-2$

7 0

2 years ago

13 please I have no idea what the steps for this is

Ilya [14]

Answer:

UV=142

Step-by-step explanation:

21x-13=10x+31

+13 +13

21x=10x+44

-10x=-10x

11x=44

÷11 ÷11

x=4

now substitute 4 in place of x (only have to solve one)

21(4)-13= 71

now multiply your answer by 2

71×2=142

6 0

3 years ago