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:
193 packets
Step-by-step explanation:
Each morning they order with a shipping fee of $10 daily.
Considering they order all 7 days of the week, so the total shipping fee for the week would be:
7 * $10 = $70
Their budget for the week is $554, out of which $70 is for shipping for the week, so remaining balance would be:
554 - 70 = $484
This 484 dollars are for coffee packets that cost $2.50 each, so the number of packets would be:
484/2.50 = 193.6
You can't order fractional packets so 193 packets is the max in this budget
Answer:
Step-by-step explanation:
3√27-10
3√9*3 -10 ( 3*9=27)
3*3√3-10( square root of 9 is 3)
<h2>9√3-10 this is simplified form</h2><h2>15.588-10=5.588 ( this answer is in decimal form)</h2>
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A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and one goes to airport C. In how many distinct ways can this be accomplished?
2.44) Refer to Exercise 2.43. Assume that taxis are allocated to airports at random.
a) If exactly one of the taxis is in need of repair, what is the probability that it is dispatched to airport C?
b) If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?
So, my answer to 2.44a is 1/9. Hopefully this is correct at least :)
For 2.44b, my guess was
(3C1)(1/3)(2/3)2 * (5C1)(1/3)(2/3)4 * 1/3
The solutions manual on chegg (which seems to be riddled with errors) says something completely different. Is my calculation correct?