Taking
and differentiating both sides with respect to
yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have
Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have
Now, when
and
, we have
Answer:
Step-by-step explanation:
You turn 5 3/5 into an improper fraction, to do that you do 5 x 5 = 25 then 25 + 3 = 28/5
so in a 7 day period she swims 5.6 kilometers. So you have to do 5.6/7 = 0.8
So it would be 0.8 kilometers in a day which is 4/5 kilometers so I don't know if the answers were just wrong?
Answer:
Step-by-step explanation:
Converse
Answer: Hello Luv......
Hope this helps. Mark me brianest please... Anna ♥
Step-by-step explanation:
Answer: -40 < T < 140
From the question, we are informed that the gas that the driver uses freezes at −40° F and evaporates at 140° F.
To get the inequality to represent the temperatures at which the gas in the truck will remain in liquid form, we should note that the temperature will be lower than 140°F but more than -40°F. This can be expressed as:
= -40 < T < 140
The answer is 6.63 Hope this helps!
