1 hour and 5 min. from 2:45 until 3:00 is 15 min. then from 3:00 until 3:50 is an additional 50 min. ad 15 and 50, and you get 65. every 60 min makes one hour, so that leaves you with one hour and five min.
Answer:
The pressure is changing at ![\frac{dP}{dt}=3.68](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3D3.68)
Step-by-step explanation:
Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.
We know that the volume is decreasing at the rate of
and we want to find at what rate is the pressure changing.
The equation that model this situation is
![PV^{1.4}=k](https://tex.z-dn.net/?f=PV%5E%7B1.4%7D%3Dk)
Differentiate both sides with respect to time t.
![\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%28PV%5E%7B1.4%7D%29%3D%20%5Cfrac%7Bd%7D%7Bdt%7Dk%5C%5C)
The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:
![\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B%7Bdx%7D%7D%5Cleft%28%20%7Bf%5Cleft%28%20x%20%5Cright%29g%5Cleft%28%20x%20%5Cright%29%7D%20%5Cright%29%20%3D%20f%5Cleft%28%20x%20%5Cright%29%5Cfrac%7Bd%7D%7B%7Bdx%7D%7Dg%5Cleft%28%20x%20%5Cright%29%20%2B%20%5Cfrac%7Bd%7D%7B%7Bdx%7D%7Df%5Cleft%28%20x%20%5Cright%29g%5Cleft%28%20x%20%5Cright%29)
Apply this rule to our expression we get
![V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0](https://tex.z-dn.net/?f=V%5E%7B1.4%7D%5Ccdot%20%5Cfrac%7BdP%7D%7Bdt%7D%2B1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%3D0)
Solve for ![\frac{dP}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D)
![V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}](https://tex.z-dn.net/?f=V%5E%7B1.4%7D%5Ccdot%20%5Cfrac%7BdP%7D%7Bdt%7D%3D-1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%5E%7B1.4%7D%7D%20%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%7D%7D)
when P = 23 kg/cm2, V = 35 cm3, and
this becomes
![\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%7D%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%2023%20%5Ccdot%20-4%7D%7B35%7D%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D3.68)
The pressure is changing at
.
Answer:
$187,244
Step-by-step explanation:
$68,000 annual salary + 3.15% commission + fee of $4.25 per transaction
$68,000 + 0.0315 * $3,600,000 + $4.25 * 1,375 =
= $187,243.75
Answer: $187,244
X² + y² - 8x - 12y + 52 = 36
x² - 8x + y² - 12y + 52 = 36
x² - 8x + y² - 12y = 88
(x² - 8x + 16) + (y² - 12y + 36) = 88 + 16 + 36
(x - 4)² + (y - 6)² = 138
(h, k) = (x, y) = (4, 6)
Answer:
B
Step-by-step explanation:
Using a graphing calculator, the graph shows an almost parabola like shape between 0 and 12
Plus if the width was equal or greater than twelve we would have a 0 or even negative volume which is impossible