Answer:
.5
Step-by-step explanation:
That 0 between 1 is .5
Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min
Answer:
Step-by-step explanation:
We are given that:
And we want to find F'(0).
First, find F(x):
![\displaystyle F'(x) = \frac{d}{dx}\left[ f(3x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20f%283x%29%5D)
From the chain rule:
![\displaystyle \begin{aligned} F'(x) &= f'(3x) \cdot \frac{d}{dx} \left[ 3x\right] \\ \\ &= 3f'(3x)\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20F%27%28x%29%20%26%3D%20f%27%283x%29%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cleft%5B%203x%5Cright%5D%20%5C%5C%20%5C%5C%20%26%3D%203f%27%283x%29%5Cend%7Baligned%7D)
Then:
In conclusion, F'(0) = 15.
12/3 or the same thing as ur question but just add a division sign, eg. difference of two and one that would be 2-1
The answers are A, C and D
