Your gonna have to put the options in the question
Answer:
A)

B)

Step-by-step explanation:
<em>x</em> and <em>y</em> are differentiable functions of <em>t, </em>and we are given the equation:

First, let's differentiate both sides of the equation with respect to <em>t</em>. So:
![\displaystyle \frac{d}{dt}\left[xy\right]=\frac{d}{dt}[6]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5Bxy%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B6%5D)
By the Product Rule and rewriting:
![\displaystyle \frac{d}{dt}[x(t)]y+x\frac{d}{dt}[y(t)]=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Bx%28t%29%5Dy%2Bx%5Cfrac%7Bd%7D%7Bdt%7D%5By%28t%29%5D%3D0)
Therefore:

A)
We want to find dy/dt when <em>x</em> = 4 and dx/dt = 11.
Using our original equation, find <em>y</em> when <em>x</em> = 4:

Therefore:

Solve for dy/dt:

B)
We want to find dx/dt when <em>x</em> = 1 and dy/dt = -9.
Again, using our original equation, find <em>y</em> when <em>x</em> = 1:

Therefore:

Solve for dx/dt:

Answer:
Step-by-step explanation:
You have no grounds for making a statement like that. There are a variety of reasons why you might not get immediate answers. Be patient.
I will do the second part of this question (finding the first three numbers):
a(4) = a(3)*(-3) + 2 = -148, so a(3)*(-3) = -150 and a(3) = -50
a(3) = 50
a(2) = a(3)*(-3) + 2 = 50, so -3*a(3) = 48 and a(d) = -16
a(1) = a(2)*(-3) + 2 = -16, so a(2)*(-3) = -18 and a(1) = 6
The procedure for finding a(5), a(6) and a(7) is exactly the same.
Answer:
3
Step-by-step explanation: